Answer:
A
Explanation:
They're easy to dislodge a difficult to move through it the crystal
Consider the balanced reversible reaction of acetic acid with ethanol, which takes place with no solvent water.
acetic acid
+
ethanol
⇌
ethyl acetate
+
water
When you react
8.29
M
acetic acid with
8.29
M
ethanol, the equilibrium concentration of acetic acid is
3.28
M
.
What is the equilibrium concentration (M) of ethyl acetate?
The equilibrium concentration of ethyl acetate is 4.02 M.
Steps
The balanced chemical equation for the reaction is:
acetic acid + ethanol ⇌ ethyl acetate + water
We are given that the initial concentration of acetic acid and ethanol is 8.29 M and that the equilibrium concentration of acetic acid is 3.28 M. Let's call the equilibrium concentration of ethyl acetate x.
Using the equilibrium constant expression for this reaction:
Kc = [ethyl acetate][water] / [acetic acid][ethanol]
We know that this reaction is balanced, meaning that the stoichiometric coefficients for the reactants and products are equal. Therefore, we can write:
Kc = [ethyl acetate][water] / [acetic acid][ethanol]
Kc = x(8.29 - 3.28) / (3.28)(8.29 - x)
Kc = 1.47
Now we can use the equilibrium constant expression to solve for x:
1.47 = x(8.29 - 3.28) / (3.28)(8.29 - x)
1.47(3.28)(8.29 - x) = x(8.29 - 3.28)
12.0437 - 1.47x = 8.29x - 27.1924
9.76x = 39.2361
x = 4.02 M
Therefore, the equilibrium concentration of ethyl acetate is 4.02 M.
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If two forces are going in opposite directions, find the net force by (A. multiply), (B. subtract), (C. add), (D. divide) the forces.
Answer:
B subtract
Explanation:
they are pulling away from one another so a and c are out of the question then that leaves you with b or d but we're not dividing anything. soooo we're left with B.
What quantity in moles of NaI are there in 175.0 mL of 0.210 M NaI?
There are 0.03675 moles of NaI in 175.0 mL of 0.210 M NaI solution.
How to calculate the number of moles ?To calculate the number of moles of NaI in 175.0 mL of 0.210 M NaI solution, we can use the formula:
moles of solute = concentration (in M) x volume (in liters)
First, we need to convert the volume from milliliters to liters:
175.0 mL = 175.0/1000 L = 0.175 L
Now we can plug in the values:
moles of NaI = 0.210 M x 0.175 L = 0.03675 moles of NaI
Therefore, there are 0.03675 moles of NaI in 175.0 mL of 0.210 M NaI solution.
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TiCl4 + 2 H2O → TiO2 + 4 HCl How many mols of TiO2 is produced with 100 g of TiCl4
0.29 mol
0.50 mol
0.98 mol
0.74 mol
0.50 mοI οf titanium diοxide is prοduced with 100 g οf titanium tetrachIοride. Thus the cοrrect answer is οptiοn (c).
Hοw dο yοu caIcuIate the number οf mοIes οf TiO2 prοduced?The baIanced chemicaI equatiοn fοr the reactiοn is:
TiCI₄ + 2 H₂O → TiO₂ + 4 HCI
Frοm the equatiοn, we can see that 1 mοIe οf TiCI₄ prοduces 1 mοIe οf TiO₂.
Tο caIcuIate the number οf mοIes οf TiO₂ prοduced frοm 100 g οf TiCI₄, we need tο first determine the number οf mοIes οf TiCI₄ present in 100 g.
The mοIar mass οf TiCI₄ is 189.68 g/mοI (47.867 + 4 x 35.453). Using this mοIar mass, we can caIcuIate the number οf mοIes οf TiCI₄ in 100 g as:
mοIes οf TiCI₄= mass οf TiCI₄/ mοIar mass οf TiCI₄
mοIes οf TiCI₄= 100 g / 189.68 g/mοI
mοIes οf TiCI₄= 0.5276 mοI
Since 1 mοIe οf TiCI₄ prοduces 1 mοIe οf TiO₂, the number οf mοIes οf TiO₂ prοduced is aIsο 0.5276 mοI.
Therefοre, the cοrrect answer is οptiοn (c) 0.50 mοI.
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Calculate the N/Z ratio for elements with atomic numbers 104 through 109. Are they in the belt of stability? Are they stable? How do you know?
The ratio of neutrons to protons, or the N/Z ratio, plays a crucial role in determining a nucleus' stability. The range of N/Z ratios in which nuclei are stable is generally referred to as the belt of stability.
How can you tell whether a substance is stable or unstable?If the forces between the constituents of the nucleus are equal, an atom is stable. If these forces are out of balance or if the nucleus has an excessive amount of internal energy, an atom is unstable (radioactive).
Z = 104 for Rutherfordium, element 104. The isotopes 261Rf and 262Rf, having masses of 261 and 262, respectively, have the longest half-lives. Accordingly, N/Z ratios are:
261Rf: N/Z = (261-104)/157 = 1.08
262Rf: N/Z = (262-104)/158 = 1.09
These N/Z ratios are a little bit higher than the average belt of stability values, which are about 1.0 for heavy nuclei. These isotopes are thought to be reasonably stable because they are close enough.
Z = 109 for Meitnerium, element 109. The isotopes 278Mt and 282Mt, with masses of 278 and 282, respectively, have the longest half-lives. Accordingly, N/Z ratios are:
278Mt: N/Z
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Can someone help me with this?
Arrhenius base - Releases OH ions when dissolved in water
Arrhenius acid - Releases H+ ions when dissolved in water
Bronsted-Lowry base - Accepts a proton
Bronsted-Lowry acid - donates a proton
How are Arrhenius bases recognized?An Arrhenius base is a molecule that decomposes into an OH- or hydroxide in solution when dissolved in water. Look for a molecule ending in OH that does not follow CHx, which denotes an alcohol, to identify the Arrhenius base. Examples of Arrhenius bases include sodium hydroxide, or NaOH.
Arrhenius acid: What is it?A substance that raises the concentration of H+ ions in an aqueous solution is known as an Arrhenius acid. Traditional Arrhenius acids are highly polarized covalent substances that dissociate in water to form an anion (A-) and the cation H+. Often, the H+ is referred to as a proton.
What distinguishes a Bronsted-Lowry base?Count the hydrogens on each component before and after the reaction to determine if it is an acid or a basic. If there are fewer hydrogens, then the substance is acid (donates hydrogen ions). The material is the base if the hydrogen count has increased (accepts hydrogen ions).
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Can someone understand this because I don’t
We need 8.8 moles of O2
We need 1.266 moles of K
We will produce 0.96 moles of water
What is a balanced reaction equation?We know that;
2H2O + O2 → 2H2O2
If 1 mole of O2 makes 2 moles of H2O2
x moles of O2 makes 17.6 moles of H2O2
x = 8.8 moles
Again;
2K + HgCl2 → 2KCl + Hg
If 2 moles of K reacts with 1 mole of HgCl2
x moles of K reacts with 0.633 moles of HgCl2
x = 1.266 moles
Again;
C3H8 + 5O2 → 3CO2 + 4H2O
5 moles of O2 produces 4 moles of H2O
1.2 moles of O2 would produce x moles of water
x = 0.96 moles
Again;
2NH3 ---->N2 + 3H2
If 1 moles of N2 is formed when 3 moles of H2 are formed
3 moles of N2 will lead to the formation of 3 * 3/1
= 9 moles of H2
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Fast please
Calculate the vapor pressure lowering, ΔP, when 10.0 mL of glycerol (C3H8O3) is added to 500 mL of water at 50oC. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.
Answer:
To calculate vapor pressure lowering:
Use ΔP = X2 * P2° equation, where X2 is the mole fraction of the solute and P2° is the vapor pressure of the pure solvent at the same temperature.
Calculate mole fraction of glycerol by using moles of glycerol and water.
Calculate vapor pressure of pure water at 50°C using the Antoine equation.
Substitute the values and calculate the vapor pressure lowering, which is 0.459 torr for this problem.
The vapor pressure lowering is 0.213 torr when 10.0 mL of glycerol is added to 500 mL of water at 50°C.
What is vapor pressure?Pressure exerted by vapor in thermodynamic equilibrium with its condensed phases at a certain given temperature in closed system is called vapor pressure.
ΔP = X2 * P0 * (1 - (ρ1 / ρ2))
ΔP is vapor pressure lowering, X2 is mole fraction of the solute (glycerol), P0 is vapor pressure of solvent (water), and ρ1 and ρ2 are the densities of solvent and solution, respectively.
As, moles of glycerol = mass of glycerol / molar mass of glycerol
moles of glycerol = (10.0 mL)(1.26 g/mL) / (92.09 g/mol)
moles of glycerol = 0.136 mol
and moles of water = mass of water / molar mass of water
= (500 mL)(0.988 g/mL) / (18.02 g/mol)
moles of water = 27.5 mol
So, total moles = moles of glycerol + moles of water
total moles = 0.136 mol + 27.5 mol
total moles = 27.6 mol
X2 (mole fraction of glycerol) = moles of glycerol / total moles
= 0.136 mol / 27.6 mol
X2 = 0.00493
ΔP = X2 * P0 * (1 - (ρ1 / ρ2))
= (0.00493)(92.5 torr) * (1 - (0.988 g/mL / 1.250 g/mL))
ΔP = 0.213 torr
Therefore, the vapor pressure lowering is 0.213 torr when 10.0 mL of glycerol is added to 500 mL of water at 50°C.
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CAN SOMEONE HELP WITH THIS QUESTION?
1234y5934y89524y5284952y542
Magnets mess touch in order to be affected by magnetic force true or false
Answer:
False. Magnets do not need to touch in order to be affected by magnetic force. Magnetic force is a non-contact force that can act on objects that are some distance apart. The strength of the force depends on the distance between the magnets and their magnetic properties.
Batrachotoxin, C31H42N2O6 , an active component of South American arrow poison, is so toxic that 0.05μg can kill a person.
In 0.05μg, there are [tex]6.9 * 10^{15} molecules[/tex] of batrachotoxin.
To calculate the number of molecules of batrachotoxin, we first need to calculate the molar mass of the molecule. This can be done by adding up the atomic masses of the elements in the compound. The elements in batrachotoxin are carbon (C), hydrogen (H), nitrogen (N), and oxygen (O). The atomic masses of these elements are 12 g/mol for C, 1 g/mol for H, 14 g/mol for N, and 16 g/mol for O. Therefore, the molar mass of batrachotoxin is:
Molar mass = 12 g/mol C + 1 g/mol H + 14 g/mol N + 16 g/mol O
Molar mass = 43 g/mol
We then need to calculate the mass of 0.05 μg of batrachotoxin. This can be done by converting 0.05 μg to grams. To do this, we divide 0.05 μg by 1,000,000. This gives us:
[tex]\frac{0.05\mu g }{ 1,000,000 }= 0.00000005 g[/tex]
Now we can calculate the number of molecules of batrachotoxin in 0.05 μg by dividing the mass in grams by the molar mass:
Number of molecules =[tex]\frac{ 0.00000005 g }{ 43 g/mol}[/tex]
Number of molecules = [tex]1.16 * 10^{-8} mol[/tex]
Finally, we can calculate the number of molecules by multiplying the number of moles by Avogadro's number:
Number of molecules =[tex]\frac{1.16 * 10^{-8 }mol * 6.022 * 10^{23 }molecules}{mol}[/tex]
Number of molecules = [tex]6.9 * 10^{15} molecules[/tex]
Therefore, there are [tex]6.9 * 10^{15} molecules[/tex] of batrachotoxin in 0.05 μg.
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complete question:Batrachotoxin, C31H42N2O6 an active component of South American arrow poison, is so toxic that 0.05μg can kill a person.
How many molecules is this? Express your answer as an integer
Hey can someone help me fill out these blanks
Answer:
Constructive interference is when two waves interact causing larger waves because the crest will meet a crest or a trough will meet a trough which adds the waves together changing the size of the amplitude.
Destructive interference is when two waves interact causing smaller waves because the crest will meet a trough or a trough will meet a crest which subtracts the waves together changing the size of the amplitude.
An ideal gas (at STP) has a volume of 5 L, how many moles of the gas are present?
Answer:
At STP (standard temperature and pressure), the conditions are:
Temperature (T) = 273.15 K Pressure (P) = 1 atm = 101.3 kPa Volume (V) = 22.4 L (for one mole of gas)
So, for a gas at STP with a volume of 5 L, we can use the following formula to calculate the number of moles present:
n = V / Vm
where: n = number of moles V = volume of gas (in liters) Vm = molar volume of gas at STP (22.4 L/mol)
Plugging in the values, we get:
n = 5 L / 22.4 L/mol n = 0.2232 mol (rounded to four significant figures)
Therefore, there are approximately 0.2232 moles of the gas present.
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Aniline, a starting compound for urethane plastic foams, consists of C, H, and N.
Combustion of such compounds yields CO2 (carbon dioxide), H2O (water), and N2 as
products. If the combustion of 9.71 mg of Aniline yields 6.63 mg of H2O and 1.46 mg of
N2, what is its empirical formula? The molecular weight of Aniline is 93 amu. What is its
molecular formula?
The empirical formula of Aniline is C9H10N and its molecular formula is C9H9.9N.
Steps
To determine the empirical formula of Aniline, we need to calculate the number of moles of each element present in the given mass of the compound and then find the smallest whole-number ratio between them.
Given:
Mass of Aniline = 9.71 mg
Mass of water produced = 6.63 mg
Mass of N2 produced = 1.46 mg
The molecular weight of Aniline = 93 amu
First, let's calculate the number of moles of water and nitrogen produced:
moles of H2O = 6.63 mg / 18.015 g/mol = 0.3680 mmol
moles of N2 = 1.46 mg / 28.014 g/mol = 0.0521 mmol
Next, we need to calculate the number of moles of carbon and nitrogen present in Aniline:
moles of C = (9.71 mg - (0.3680 mmol x 12.011 g/mol)) / 12.011 g/mol = 0.4811 mmol
moles of N = 0.0521 mmol
Now, we need to find the smallest whole-number ratio between these elements. We can divide the number of moles of each element by the smallest value, which is 0.0521 mmol:
moles of C = 0.4811 mmol / 0.0521 mmol = 9.231 ≈ 9
moles of N = 1
moles of H = not calculated, but we can find it using the difference in mass between Aniline and the products formed.
Now, let's calculate the mass of carbon and nitrogen in Aniline:
mass of C = 9 x 12.011 g/mol = 108.099 g/mol
mass of N = 1 x 14.007 g/mol = 14.007 g/mol
Finally, we can calculate the mass of hydrogen by taking the difference between the mass of Aniline and the sum of the masses of carbon, nitrogen, water, and nitrogen:
mass of H = 9.71 mg - (108.099 g/mol + 14.007 g/mol + 6.63 mg + 1.46 mg) = 2.47 mg
Now, we can calculate the number of moles of hydrogen:
moles of H = 2.47 mg / 1.008 g/mol = 2.449 mmol
Finally, we can express the empirical formula of Aniline as: C9H10N
To find the molecular formula, we need to calculate the molecular weight of the empirical formula:
Empirical formula weight = (9 x 12.011 g/mol) + (10 x 1.008 g/mol) + (1 x 14.007 g/mol) = 93.126 g/mol
Now, we can find the molecular formula by dividing the molecular weight of Aniline by the empirical formula weight and multiplying each subscript in the empirical formula by the result:
The molecular weight of Aniline = 93 amu
The molecular weight of the empirical formula = 93.126 g/mol
Molecular formula = empirical formula x (Molecular weight of Aniline / Empirical formula weight)
= C9H10N x (93 amu / 93.126 g/mol)
= C9H9.9N
Therefore, the empirical formula of Aniline is C9H10N and its molecular formula is C9H9.9N.
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What is the empirical formula for a compound containing 37.5% carbon, 12.6% hydrogen, and 49.9% oxygen? A. CH₂O B. C₂HO5 C. C₂H₁203 D. C₂H₂O₂
The empirical formula for the compound containing 37.5% carbon, 12.6% hydrogen, and 49.9% oxygen is CH₄O
How do i determine the empirical formula?The following data were obtained from the question:
Carbon (C) = 37.5%Hydrogen (H) = 12.6%Oxygen (O) = 49.9%Empirical formula =?From the above data, we can obtain the empirical formula for the compound as shown below:
Divide by their molar mass
C = 37.5 / 12 = 3.125
H = 12.6 / 1 = 12.6
O = 49.9 / 16 = 3.119
Divide by the smallest
C = 3.125 / 3.119 = 1
H = 12.6 / 3.119 = 4
O = 3.119 / 3.119 = 1
Thus, we can conclude that the empirical formula is CH₄O
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These two samples of NaCl and CoCl2 have the same number of moles. Do they have the same mass? Do they have the same number of particles? Explain.
No, two samples of NaCl and CoCl2 do not have the same mass, but they have the same number of particles.
What is meant by moles?Mole is the amount of material containing 6.02214 × 10²³ particles.
Molar mass of NaCl (sodium chloride) is 58.44 g/mol, while molar mass of CoCl2 (cobalt chloride) is 129.84 g/mol. Since both samples have same number of moles, it means that they contain same number of particles of their respective compounds. However, mass of each sample will be different due to the difference in molar mass.
For example, if we assume that both samples contain 1 mole of their respective compounds, then mass of NaCl sample will be 58.44 g, while mass of CoCl2 sample will be 129.84 g. So, two samples have different masses but the same number of particles.
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Calculate the change in pH when 2.0×10−2mol of NaOH is added to 0.50 L of a buffer solution that is 0.15 M in HF and 0.20 M in NaF .
Answer:
The change in pH is 0.04.
how many significant figure are in 0.03412
The measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.
What is measurements?Measurements are the numerical values used to quantify the size, amount, or extent of something. Measurements are used to describe physical attributes and characteristics, such as length, width, weight, height, area, volume, speed, temperature, pressure, force, energy, power, and time. Measurements are also used to determine the amount of a certain material that is needed for a given purpose, such as a recipe. Measurements are essential for scientific investigation, engineering, and other activities that require precision and accuracy.
The measurement of 0.03412 kg has 5 significant figures. Significant figures are the meaningful digits of a number that are important in providing information about the measurement. In this measurement, the 4 digits after the decimal point (3412) provide the precision of the measurement, and therefore are significant figures. The leading zero does not provide any additional precision and is not a significant figure. Therefore, the measurement of 0.03412 kg has 5 significant figures.
Significant figures are used to provide an indication of the precision of a measurement. For example, a measurement of 0.03412 kg has a precision of 0.00001 kg, which is greater than a measurement of 0.3412 kg, which has a precision of 0.0001 kg. The more significant figures a measurement has, the more precise it is.
Significant figures are also important in calculations. To ensure accuracy, all calculations should be done using the same number of significant figures as the measurements being used. For example, if two measurements, 0.03412 kg and 0.02384 kg, are used in a calculation, the result should be reported to 5 significant figures.
In conclusion, the measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.
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The measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.
What is measurements?Measurements are the numerical values used to quantify the size, amount, or extent of something. Measurements are used to describe physical attributes and characteristics, such as length, width, weight, height, area, volume, speed, temperature, pressure, force, energy, power, and time. Measurements are also used to determine the amount of a certain material that is needed for a given purpose, such as a recipe. Measurements are essential for scientific investigation, engineering, and other activities that require precision and accuracy.
The measurement of 0.03412 kg has 5 significant figures. Significant figures are the meaningful digits of a number that are important in providing information about the measurement. In this measurement, the 4 digits after the decimal point (3412) provide the precision of the measurement, and therefore are significant figures. The leading zero does not provide any additional precision and is not a significant figure. Therefore, the measurement of 0.03412 kg has 5 significant figures.
Significant figures are also important in calculations. To ensure accuracy, all calculations should be done using the same number of significant figures as the measurements being used. For example, if two measurements, 0.03412 kg and 0.02384 kg, are used in a calculation, the result should be reported to 5 significant figures.
In conclusion, the measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.
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The complete question is as follows:
How many significant figure are in 0.03412 kg?
Calculate the concentration of an aqueous solution of Ca(OH)2
that has a pH
of 11.67.
Ca(OH)2 in an aqueous solution with a pH of 11.67 has a concentration of 2.50 x 10(-3) M.
What is the pH based on the Ca OH 2 concentration?The calcium hydroxide aqueous solution has a pH of 11.03. Two moles of hydroxide ions are created from one mole of an aqueous solution of calcium hydroxide.
The pH of a solution can be related to the concentration of hydroxide ions ([OH-]) through the equation: pH + pOH = 14
We can rearrange this equation to solve for [OH-]: [OH-] = 10^(-pOH)
Thus: Ca(OH)2 → Ca^2+ + 2 OH-
Since the molar ratio of Ca(OH)2 to [OH-] is 1:2, the concentration of hydroxide ions in the saturated solution can be calculated as follows:
[OH-] = 2 x [Ca(OH)2]
Now we can use the pH value given in the problem to calculate the pOH:
pOH = 14 - pH
pOH = 14 - 11.67
pOH = 2.33
Substituting this value into the equation for [OH-]:
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.33)
[OH-] = 5.01 x 10^(-3) M
In order to get the concentration of the solution, we can apply the equation for the concentration of hydroxide ions in a saturated solution of Ca(OH)2:
[Ca(OH)2] = [OH-] / 2
[Ca(OH)2] = 5.01 x 10^(-3) M / 2
[Ca(OH)2] = 2.50 x 10^(-3) M
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oxidation take place at the anode during electrolysis because the anode is (a) is deficient in electrode (B)is deficient in proton (c) has excess electron (d)has attraction for positive ion
Answer : Option (c) is correct.
During electrolysis, a process in which an electric current is passed through a solution containing ions, two electrodes are immersed in the electrolyte solution. The cathode is the negatively charged electrode, while the anode is the positively charged electrode. The electrolyte solution contains both positively charged ions (cations) and negatively charged ions (anions).
At the anode, oxidation occurs as the positively charged ions (cations) in the electrolyte are attracted to the negatively charged anode. The cations lose electrons and become neutral atoms, and these electrons are transferred to the anode. This loss of electrons by the cations results in their oxidation.
The anode, being the electrode where oxidation occurs, has an excess of electrons, which are attracted by the positively charged cations in the electrolyte. The excess of electrons is due to the fact that the anode is connected to the positive terminal of the power source, which supplies electrons to the electrode.
moles of Zn(NO 3 ) 2 in 173.50 g of this substances
Answer:
To calculate the moles of Zn(NO3)2 in 173.50 g of the substance, we first need to know the molar mass of Zn(NO3)2.
The molar mass of Zn(NO3)2 can be calculated by adding the atomic masses of its constituent elements:
Zn: 1 x 65.38 g/mol = 65.38 g/mol
N: 2 x 14.01 g/mol = 28.02 g/mol
O: 6 x 16.00 g/mol = 96.00 g/mol
Molar mass of Zn(NO3)2 = 65.38 + 28.02 + 96.00 = 189.40 g/mol
Now we can use this molar mass to calculate the moles of Zn(NO3)2 in 173.50 g of the substance:
moles = mass/molar mass
moles = 173.50 g/189.40 g/mol
moles = 0.9164 mol
Therefore, there are 0.9164 moles of Zn(NO3)2 in 173.50 g of the substance.
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Identify the precipitation
Precipitation reactions and non- precipitation reactions
When two aqueous solutions are combined, a precipitation process takes place in which an insoluble substance (precipitate) develops. Reactions 1, 2, 3, and 5 in the list are precipitation reactions.
What are some examples of precipitation reactions?When an impermeable material called a precipitate separates from the solution, a reaction known as a precipitation reaction takes place. As an example, a white precipitate of barium sulphate and sodium chloride solution is created when sodium sulphate solution and barium chloride solution are mixed.
Describe precipitates and provide an example.An insoluble material is called a precipitate. For instance, barium sulphate and sodium chloride solution is created when sodium sulphate solution and barium chloride solution are combined.
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Determine the empirical formula of a compound that is composed of 39.27g Iron and 33.77g sulfer (please give indepth explanation)
Answer:
To determine the empirical formula of a compound, we need to know the relative amounts of each element in the compound. Given the masses of iron and sulfur, we can calculate the number of moles of each element present:
moles of iron = 39.27 g / 55.85 g/mol = 0.703 mol
moles of sulfur = 33.77 g / 32.06 g/mol = 1.053 mol
We can then find the ratio of these moles by dividing both by the smallest number of moles (0.703):
0.703 mol Fe : 1.053 mol S
0.667 Fe : 1.000 S
This ratio indicates that there are approximately 0.667 atoms of iron for every 1 atom of sulfur in the compound.
To convert this ratio to a whole-number ratio of atoms, we need to multiply by a factor that will give us whole numbers. We can do this by dividing both sides of the ratio by the smallest number of atoms:
0.667 Fe : 1.000 S
0.667/0.667 Fe : 1.000/0.667 S
1.000 Fe : 1.498 S
This gives us a ratio of approximately 1 atom of iron for every 1.498 atoms of sulfur. To get a whole-number ratio, we can multiply both sides by 2:
2.000 Fe : 2.996 S
Rounding to the nearest whole number, we get a ratio of:
2 Fe : 3 S
Therefore, the empirical formula of the compound is Fe2S3.
the presence of chloride ions in a sample can be detected by reacting it with silver nitrate, agno3. if agno3 is added to an aqueous solution of sodium chloride; silver chloride, which contains cl- ions, is essentially insoluble in water, will precipitate from solution as a white solid. create a balanced equation of this reaction.
The balanced equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) to form silver chloride (AgCl) and sodium nitrate (NaNO3) is:
AgNO3 + NaCl → AgCl + NaNO3
A balanced equation is a representation of a chemical reaction that shows the reactants and products involved, as well as the ratios in which they combine. The law of conservation of mass states that in a chemical reaction, the mass of the reactants must be equal to the mass of the products, which means that the number and type of atoms on both sides of the equation must be the same.
To balance an equation, coefficients are added to the reactants and products so that the number of atoms of each element on the left side of the equation is equal to the number on the right side. For example, the combustion of methane gas can be represented by the equation CH4 + 2O2 → CO2 + 2H2O. Balanced equations are essential for understanding chemical reactions and predicting the amount of reactants needed and products formed.
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What are some ways you can increase the percent yield of this reaction?
If the reaction's measured product has impurities that increase its mass over what it would be if it were pure, higher percent yields than 100% are attainable.
What influences the increase or decrease in % yield?Because the real yield is frequently lower than the theoretical value, percent yield is typically lower than 100%. This may be due to incomplete or conflicting reactions or sample loss during recovery.
What does an increase in yield mean?It implies that interest rates will increase further, yields will increase, and bond prices would decline as a result. So, it is likely that the bond market will continue to see unusually high levels of volatility in the near future.
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2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g) in word form
Two solid potassium (K) combine with two liquid water (H2O) molecules to generate two aqueous potassium hydroxide (KOH) and one gaseous hydrogen (H2) molecule. For this reaction, the chemical equation is balanced as follows:
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
Steps
The chemical reaction between potassium and water is depicted in this equation. Two liquid water (H2O) molecules and two solid potassium (K) atoms serve as the reactants.
The reaction moves from the left to the right, as shown by the arrow sign, which also denotes its direction.
In the reaction, hydrogen gas and potassium hydroxide are created when potassium atoms interact with water molecules.
Two molecules of aqueous potassium hydroxide (KOH) and one molecule of gaseous hydrogen (H2) are the reaction's end products.
According to the correctly balanced chemical equation, two potassium atoms and two water molecules combine to form two molecules of potassium hydroxide and one hydrogen gas molecule.
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Question 6 of 10
Which prefix indicates a molecule with 7 carbon atoms?
OA. Non-
OB. Dec-
C. Hept-
OD. Eth-
Answer:
Hept
Explanation:
Non- 9 C9
Dec- 10 C10
Eth- 2 C2
Hepth- 7 C7
So the answer is D, because it idicates a molecule with 7 carbon atoms.
Hopefully this helps! :)
Chemistry Help Please! It's worth a lot of points
1.Copper is commonly used to make electrical wires. How many moles of copper are in 5.00 grams of copper wire?
2.Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has a molecular formula of C2H5O2N. How many moles of glycine molecules are contained in 28.35 grams of glycine?
3. Vitamin C is a covalent compound with the formula of C6H8O6. The recommended daily dietary allowance of vitamin C for children aged 4-8 years is 1.42 x 10-4 mol.
a. What is the mass of this allowance in grams?
b. How many moles of carbon are in 1.42 x 10-4 mol of C6H8O6?
Answer:
1. To determine the number of moles of copper in 5.00 grams of copper wire, we need to use the molar mass of copper. The molar mass of copper is 63.55 g/mol. We can use the following conversion factor:
1 mol Cu = 63.55 g Cu
Using this conversion factor, we can calculate the number of moles of copper:
5.00 g Cu × (1 mol Cu / 63.55 g Cu) = 0.0787 mol Cu
Therefore, there are 0.0787 moles of copper in 5.00 grams of copper wire.
2. To determine the number of moles of glycine molecules in 28.35 grams of glycine, we need to use the molar mass of glycine. The molar mass of glycine is 75.07 g/mol. We can use the following conversion factor:
1 mol glycine = 75.07 g glycine
Using this conversion factor, we can calculate the number of moles of glycine molecules:
28.35 g glycine × (1 mol glycine / 75.07 g glycine) = 0.3778 mol glycine
Therefore, there are 0.3778 moles of glycine molecules in 28.35 grams of glycine.
3. a. To determine the mass of the daily dietary allowance of vitamin C in grams, we can use the following conversion factor:
1 mol C6H8O6 = 176.12 g C6H8O6
Using this conversion factor, we can calculate the mass of the allowance:
1.42 × 10^-4 mol C6H8O6 × (176.12 g C6H8O6 / 1 mol C6H8O6) = 0.0248 g
Therefore, the mass of the daily dietary allowance of vitamin C for children aged 4-8 years is 0.0248 grams.
b. To determine the number of moles of carbon in 1.42 × 10^-4 mol of C6H8O6, we can use the molar mass of carbon. The molar mass of carbon is 12.01 g/mol. There are 6 carbons in each molecule of C6H8O6, so we can use the following conversion factor:
6 mol C / 1 mol C6H8O6
Using this conversion factor, we can calculate the number of moles of carbon:
1.42 × 10^-4 mol C6H8O6 × (6 mol C / 1 mol C6H8O6) = 8.52 × 10^-4 mol C
Therefore, there are 8.52 × 10^-4 moles of carbon in 1.42 × 10^-4 mol of C6H8O6.
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CuCl2(aq)+Na2CO3(aq) complete and balance the precipitation reaction.
Answer: Na2CO3(aq) + CuCl2(aq) → 2 NaCl(aq) + CuCO3(s).
Explanation:
use dimensional analysis to solve all of the following:
0.5 moles of [tex]H_{2[/tex][tex]O_{2}[/tex] produces 8 grams of [tex]O_{2}[/tex].
What is Moles?
Mole is a unit of measurement used in chemistry to express the amount of a substance. It is defined as the amount of a substance that contains the same number of entities (such as atoms, molecules, or ions) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number.
To use dimensional analysis, we need to set up the given equation in terms of units. We can use the molar mass of[tex]H_{2}[/tex][tex]O_{2}[/tex] and [tex]O_{2}[/tex] to convert between moles and grams:
2[tex]H_{2}[/tex]O2 → 2[tex]H_{2}[/tex] + [tex]O_{2}[/tex]
The balanced equation shows that 2 moles of [tex]H_{2}[/tex][tex]O_{2}[/tex] produce 1 mole of [tex]O_{2}[/tex]. We can use this ratio to convert between moles of [tex]H_{2}[/tex][tex]O_{2}[/tex]and moles of[tex]O_{2}[/tex].
0.5 moles of [tex]H_{2}[/tex]O2 x (1 mole of [tex]O_{2}[/tex] / 2 moles of[tex]H_{2}[/tex][tex]O_{2}[/tex]) = 0.25 moles of [tex]O_{2}[/tex]
Now we can use the molar mass of [tex]O_{2}[/tex] to convert moles to grams:
0.25 moles of [tex]O_{2}[/tex] x (32 g of [tex]O_{2}[/tex] / 1 mole of [tex]O_{2}[/tex]) = 8 g of [tex]O_{2}[/tex]
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