If a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr, the molar mass of the compound is 24.8 g/mol.
To calculate the molar mass of the compound, we first need to calculate the number of moles present in the gaseous sample using the ideal gas law:
PV = nRT
Where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
Converting the given pressure of 615 torr to atm:
615 torr = 0.811 atm
Converting the given temperature of 30°C to Kelvin:
30°C + 273.15 = 303.15 K
Rounding off to two significant figures, we get:
P = 0.81 atm
T = 303 K
Now, rearranging the ideal gas law equation to solve for n:
n = PV/RT
Substituting the given values:
n = (0.978 g/L) x (1 L) / (0.081 atm x 0.0821 L atm/mol K x 303 K)
n = 0.0394 mol
Next, we can calculate the molar mass of the compound using the formula:
molar mass = mass / mole
molar mass = (0.978 g/L) x (1 L) / 0.0394 mol
molar mass = 24.8 g/mol
Therefore, 24.8 g/mol is the molar mass of the compound.
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Hellpppp with this question!!! THE ANSWER IS NOT 0.3 or 0.5
the answer is 2.5 according to me
Calculate the value of Kp at 227 degrees Celsius for the equilibrium: 3 A(g) ⇌ B(g) + D(g Kc=5.15
Which
thermochemical
equation
corresponds to
the graph?
Answer: C
Explanation:
Answer: C
Explanation:
What is the molarity of a solution that has 2.0 moles of solute in 3.0 L of solution?
The molarity of the solution that has 2.0 moles of solute in 3.0 L of solution is 0.67 mol/L
What is molarity?Molarity is described as a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.
Molarity = moles of solute / liters of solution
we then substitute the given values, and have
Molarity = 2.0 moles / 3.0 L
Molarity = 0.67 mol/L
Molarity is very important because the ration used to express the concentration of any solution.
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Calculate standard cell potential of an electrochemical cell powered by these half-reactions. (Write values to two decimal places. If a value is less than 1, be sure to write a 0 before the decimal.)
Pb4+ + 2e− → Pb2+
Co3+ + e− → Co2+
E°cell = V
Is the reaction spontaneous
The standard cell potential is found as +1.95 V and is a spontaneous reaction.
What is standard cell potential ?The standard cell potential (E°cell) of an electrochemical cell is given by the difference between the standard reduction potentials of the two half-cells involved.
E°cell = E°reduction (cathode) - E°reduction (anode)
The half-reactions given are:
Pb4+ + 2e− → Pb2+ (reduction)
Co3+ + e− → Co2+ (reduction)
The standard reduction potentials for these half-reactions are:
E°reduction(Pb4+/Pb2+) = -0.13 V
E°reduction(Co3+/Co2+) = +1.82 V
We then calculate as:
E°cell = E°reduction (Co3+/Co2+) - E°reduction (Pb4+/Pb2+)
E°cell = (+1.82 V) - (-0.13 V)
E°cell = +1.95 V
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What is the number of molecules of NO, which contains 16 gm of oxygen. 14
A 210.00 g sample of water with an initial temperature of 29.0°C absorbs 7,000.0 J of heat. What is the final temperature of the water?
Note: Use C (capital C) for degrees Celsius when typing units. So it might look like 35C or 2.03 J/gC. Give your answer in 3 sig figs.
The 210.00 g sample of the water with the initial temperature of the 29.0°C absorbs the 7,000.0 J of heat. The final temperature of the water is the 36.9 °C .
The mass of the water = 210 g
The initial temperature = 29.0 °C
The final temperature = ?
The heat energy = 7000 J
The specific heat capacity = 4.184 J/g °C
The heat energy is expressed as :
Q = m c ΔT
Where,
The m is mass of water = 210 g
The c is specific heat of water = 4.184 J/g °C
The ΔT is change in temperature = final temperature - initial temperature
The ΔT is change in temperature = T - 29.0 °C
7000 = 210 × 4.184 ( T - 29.0 )
7000 = 878.64 ( T - 29.0 )
( T - 29.0 ) = 7.966
T = 36.9 °C
The final temperature is 36.9 °C .
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2ch4 and c2h8 how are they different
Answer:
Explanation:
Both 2CH4 and C2H8 have the same number and kind of elements. But practically, 2CH4 will be existing but C2H8 cannot exist.
8. The compound C2H4 has van der Waals constants a = 4.612 atm•L2/mol2 and b = 0.0582 L/mol. Using both the ideal gas law and van der Waals’s equation, calculate the pressure expected for 30 mol of C2H4 gas in a 6.00-L container at 20 °C.
Using the Ideal Gas Law, the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.
To calculate the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C, we will use both the Ideal Gas Law and van der Waals equation.
Ideal Gas Law: PV = nRT
P = pressure
V = volume (6.00 L)
n = moles (30 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (20 °C + 273.15 = 293.15 K)
Solve for P (pressure):
P = nRT / V
P = (30 mol)(0.0821 L•atm/mol•K)(293.15 K) / 6.00 L
P = 1210.07 atm
Van der Waals equation:
(P + a(n/V)²)(V - nb) = nRT
a = 4.612 atm•L²/mol²
b = 0.0582 L/mol
Solve for P (pressure):
(P + (4.612)(30/6)²) (6 - 0.0582 * 30) = (30)(0.0821)(293.15)
P = 1179.71 atm
Using the Ideal Gas Law, the pressure is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.
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What is volume of 12.0 g of carbon dioxide at stp?
Answer: 6.11 L
Explanation:
STP= 1atm, 273.15K
Molar mass of CO2=44.01g/mol so n= (12.0/44.01)
PV=nRT
V=(nRT)/P
V=((12.0/44.01)(0.0821)(273.15))/1
V=6.11L
Calculate the mass of Kr
in a 9.95 L
cylinder at 91.2 ∘C
and 4.50 bar
.
Help please! I'll give brainliest and 5 stars if you show work!
To solve this problem, we can use the formula:
q = m × c × ΔT
where q is the heat absorbed or released, m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature.
First, let's calculate the mass of water:
m = 225.0 g
Next, let's calculate the heat absorbed by the water:
q_water = m × c × ΔT
q_water = 225.0 g × 4.184 J/(g·°C) × (24.60°C - 20.53°C)
q_water = 3749.8 J
Since the metal released 4274 J of heat, the heat absorbed by the calorimeter can be calculated by subtracting the heat absorbed by the water from the total heat released by the metal:
q_calorimeter = - (q_water + q_metal)
q_calorimeter = - (3749.8 J + 4274 J)
q_calorimeter = - 8023.8 J
Therefore, the heat absorbed by the calorimeter is -8023.8 J, which is approximately equal to -8000 J or -8.0 kJ. The answer is (c) -339 J, since it is the closest to the calculated value when rounded to the nearest integer. Note that the negative sign indicates that the calorimeter absorbed the heat, which is expected since the reaction involved a release of heat.
How many grams in 5 moles of water?
Answer:
90g
Explanation:
Ans. 90 gram
we know that,
n = wt/m.wt
where, n= moles
wt.= weight
m.wt = molecular weight
putting values we get
5 = wt./18 ( molecular weight of water is 18
wt.= 90
hence ans.= 90 gram
please help asap!
3. A double replacement reaction occurs between two solutions of lead (II) nitrate and potassium bromide. Write a
balanced equation for this reaction-identifying the product that will precipitate, and the product that will remain in
solution.
a) Write the balanced equation for this double replacement reaction.
b) If this reaction starts with 32.5 g lead (II) nitrate and 38.75 g potassium bromide, how many grams of the
precipitate will be produced? Remember to use the limiting reactant to calculate the amount of precipitate
formed.
c) How many grams of the excess reactant will remain?
Answer:
Explanation:
a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:
Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)
In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.
b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.
The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.
The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol
According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.
Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.
The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.
The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.
c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.
The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.
The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.
The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.
Arrange the following ions in order of increasing ionic radius: selenide ion, rubidium ion, bromide ion, strontium ion.
Answer:
Br, Se, Sr, Rb
Explanation:
Atomic radius increases as you move to the left and down the periodic table. The increase in radius as you move left is due to decreasing effective nuclear charge (the pull an electron feels from the nucleus) since the number of protons decrease. The increase in radius as you move down is due to a higher number of principle energy levels (orbital in which the electron is located relative to the atom's nucleus), causing the electrons to be farther from the nucleus.
You react 0.017 mol of solid metal with HCl in a coffee cup calorimeter (reaction shown below). The calorimeter has 100 mL of water in it, and the temperature of the water increases by 3.81°C. The calorimeter has a heat capacity of 40.4 J/°C. What is the enthalpy of the reaction in terms of kJ per mol of the metal (your answer should be NEGATIVE, remember to convert from J to kJ, specific heat capacity of water is 4.184 J/g-°C)?
M(s) + 2 HCl (aq) MCl2 (aq) + H2 (g)
M = metal
If the reaction A (aq) + B (aq) C(aq) has a Ka value equal to 4.26 x 10-6, what is the G value at 25 °C if the concentrations are as follows:
[A] = 1.50 M
[B] = 1.00 M
[C] = 5.00 x 10-5 M
The ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.
The relationship between ΔG and K is given by the following equation:
ΔG = -RTln(K)
where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 °C = 298.15 K), and ln denotes the natural logarithm.
To calculate K, we need to use the equilibrium expression and the given concentrations:
[tex]K = [C]/([A][B])[/tex]
[tex]K = (5.00 * 10^{-5} M)/((1.50 M)(1.00 M))[/tex]
[tex]K = 3.33 x 10^{-5}[/tex]
Now we can substitute the values for R, T, and K into the equation for ΔG:
ΔG = -RTln(K)
ΔG = [tex]-(8.314 J/(mol.K))(298.15 K)ln(3.33 x 10^{-5})[/tex]
ΔG = -8.35 kJ/mol
Therefore, the ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.
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Write the electronic configuration of all the metal ions in the d-blocks (3d series)
The electronic configuration of the d-block metal ions in the 3d series is represented by electronic configuration of Argon (Ar), 3d and 4s sub orbitals.
What is the electronic configuration of all d block?The electronic configuration of the d-block metal ions in the 3d series is as follows:
Scandium (Sc): [Ar] 3d¹ 4s²
Titanium (Ti): [Ar] 3d² 4s²
Vanadium (V): [Ar] 3d³ 4s²
Chromium (Cr): [Ar] 3d⁵ 4s¹
Manganese (Mn): [Ar] 3d⁵ 4s²
Iron (Fe): [Ar] 3d⁶ 4s²
Cobalt (Co): [Ar] 3d⁷ 4s²
Nickel (Ni): [Ar] 3d⁸ 4s²
Copper (Cu): [Ar] 3d¹⁰ 4s¹
Zinc (Zn): [Ar] 3d¹⁰ 4s²
Thus, the above illustration shows the electronic configuration of all the metal ions in the d-blocks (3d series).
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A sample of an ideal gas has a volume of 2.31 L
at 279 K
and 1.01 atm.
Calculate the pressure when the volume is 1.09 L
and the temperature is 308 K.
We can use the combined gas law to determine the pressure of the gas at the final state. The combined gas law relates the pressure, volume, and temperature of a gas:
(P1 x V1) / T1 = (P2 x V2) / T2
where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.
We are given the initial pressure (P1 = 1.01 atm), volume (V1 = 2.31 L), and temperature (T1 = 279 K) of the gas, and the final volume (V2 = 1.09 L), and temperature (T2 = 308 K) of the gas. We can solve for P2, the final pressure of the gas:
(P1 x V1) / T1 = (P2 x V2) / T2
P2 = (P1 x V1 x T2) / (V2 x T1)
P2 = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)
P2 = 2.41 atm (rounded to three significant figures)
Therefore, the pressure of the gas when the volume is 1.09 L and the temperature is 308 K is approximately 2.41 atm.
WHEN SOME PEOPLE HAVE AN UPSET STOMACH, THEY TAKE A SODA TABLET LIKE
TUMS TO NEUTRALIZE THEIR STOMACH ACID.
THE REACTION IS HYDROCHLORIC ACID PLUS SODIUM BICARBONATE MAKES SALT,
CARBON DIOXIDE (THAT'S WHY SOME PEOPLE BURP) AND WATER.
HOW MUCH CARBON DIOXIDE AND SALT (IN GRAMS) ARE PRODUCED IF A 2 GRAM
TABLET OF SODIUM BICARBONATE IS TAKEN TO REACT WITH 18 GRAMS OF
HYDROCHLORIC ACID?
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium bicarbonate [tex](NaHCO_3)[/tex] is:
[tex]HCl + NaHCO_3\ - > NaCl + CO_2 + H_2O[/tex]
The coefficients in the balanced equation show that 1 mole of HCl reacts with 1 mole of [tex]NaHCO_3[/tex] to produce 1 mole of NaCl, 1 mole of [tex]CO_2[/tex], and 1 mole of [tex]H_2O[/tex]. We need to find the number of moles of [tex](NaHCO_3)[/tex] present in the tablet.
2 grams of [tex]NaHCO_3[/tex] is equivalent to 0.02 moles, and 18 grams of HCl is equivalent to 0.45 moles. Since [tex](NaHCO_3)[/tex] is limiting reagent, only 0.02 moles of NaCl and [tex]CO_2[/tex] will be produced. The molar mass of [tex]CO_2[/tex] is 44 g/mol, so the mass of [tex]CO_2[/tex] produced is 0.88 g. The molar mass of NaCl is 58.44 g/mol, mass of NaCl produced is 1.17 g.
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Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
The Correct answer is option 3
Step by Syep Explanation:
4Fe+3O2---->rust
formula for rust----->Fe2O3
4Fe+3O2---->Fe2O3
Balancing the Chemical Equation
both the reactant and product side
we have that;
4Fe+3O2------->2FeO3
the equation is Chemically Balanced
therefore 4Fe+3O2------->2×rust
Answer:
2Fe₂O₃ (Option 3)Explanation:
Given that,
4Fe + 3O2 → rust.Law of conservation of mass states that " Mass of reactants is equal to the mass of products".
Also we know that In a balanced equation the total number of atoms in the reactants equals the total number of atoms in the product.
We are given with 4Fe + 3O₂ i.e the reactant.
First Let's calculate the number of atoms in the reactant.
No. of atoms in Fe = 4 No. of atoms in O = 3 × 2 = 6Now, Let's find the product .
Also, We can see 2Fe₂O₃ (Product)
No. of atoms in Fe = 2 × 2 = 4 No. of atoms in O = 2 × 3 = 6.4Fe + 3O₂ → 2Fe₂O₃
Number of atoms in the reactants = the total number of atoms in the product.
Therefore, 2Fe₂O₃ (Option 3) will the required answer .
Please ASAP!! :'(
Which of the following graphs repMagnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial.
· Trial 1:
· Trial 2:
Determine the percent yield of MgO for your experiment for each trial.
· Trial 1:
· Trial 2:
Determine the average percent yield of MgO for the two trials.
resents the function g (x) = x2(x + 1)(x – 2)?
The theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.
To calculate the theoretical yield of MgO, we need to use the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO):
2Mg + O₂ → 2MgO
According to the stoichiometry of this equation, 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, we need to determine the number of moles of Mg in each trial and use the mole ratio to find the theoretical yield of MgO.
For Trial 1:
The mass of Mg used is: 26.682 g - 27.012 g = 0.330 g
The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg is:
0.330 g / 24.31 g/mol = 0.0136 mol Mg
According to the balanced equation, 2 moles of Mg produce 2 moles of MgO, so the theoretical yield of MgO is:
0.0136 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.348 g MgO
For Trial 2:
The mass of Mg used is: 26.987 g - 26.695 g = 0.292 g
The number of moles of Mg is:
0.292 g / 24.31 g/mol = 0.0120 mol Mg
The theoretical yield of MgO is:
0.0120 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.307 g MgO
To calculate the percent yield of MgO, we need to use the following formula:
Percent yield = (actual yield / theoretical yield) x 100%
For Trial 1:
The actual yield of MgO is: 27.214 g - 27.012 g = 0.202 g MgO
The percent yield of MgO is:
(0.202 g / 0.348 g) x 100% = 58.0%
For Trial 2:
The actual yield of MgO is: 27.183 g - 26.695 g = 0.488 g MgO
The percent yield of MgO is:
(0.488 g / 0.307 g) x 100% = 159.2%
To calculate the average percent yield of MgO for the two trials, we add the percent yields and divide by 2:
Average percent yield = (58.0% + 159.2%) / 2 = 108.6%
Therefore, the theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.
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please help me with this lab i wasn’t here for!
3. now that you have the mass of the NaHCO3 reactant, and the mass of the product NaCI , convert each to moles and compare to the mole ratio from your balanced equation C space below for your calculations
mass NaHCO3:
mass NaCI:
moles NaHCO3:
moles NaCI:
does the mole to mole ratio for your reaction? Agree with the ratio for the balanced equation?___
4. which reactant is the excess reactant for your reaction, how do you know?
5. Using the limiting reactant calculate the maximum amount of product that can be made from this reaction.
6. using the theoretical yield in the mass of the product that you put produce calculate percent yield.
calculations:
question #3: converting mass to moles
question #5: calculating the theoretical yield
question #6: calculating percent yield
Question #3: 0.8 g of NaHCO3 mass NaCI weight: 0.4 g 0.8 g/84 g/mol, or 0.0095 moles, of NaHCO3 0.4 g/58.5 g/mol = 0.0068 moles of NaCI are the moles.
The reaction's mole to mole ratio and the ratio in the balanced equation (1:1) are in agreement. The highest quantity of NaCI that may be produced from this reaction is 0.0095 moles since NaHCO3 is the limiting reactant.
The theoretical yield of NaCI is 0.0095 moles, which is question #6. The finished product weighs 0.4 g. The percent yield is 0.4 g/0.0095 moles times 100, which is 42.1%.
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In the titration between hcl and naoh what’s the medium at the end point and why ?
In the titration between HCl and NaOH, the medium is neutral at the end point because of complete neutralization of a strong acid by a strong base.
Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.
The strong acid and strong base neutralization have a pH value of 7.
The beaker gets warm which indicates that the reaction between acid and base is an exothermic reaction releasing heat energy into the surroundings.
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aHow does the electronic configuration of a sodium cation differ from that of a sodium atom?
Answer:
Atomic number of sodium is 11
Electronic configuration of a sodium atom :
1s² 2s² 2p⁶ 3s¹Since sodium has one electron in its outermost shell, Therefore, sodium can easily donate it's one electron. As the result it becomes sodium cation with + 1 charge.
Electronic configuration of a sodium cation,[tex] \: \sf ({Na}^{+1}) [/tex]
1s² 2s² 2p⁶In case of sodium cation, it has fully filled electronic configuration.
Cations - Atoms that carry postive charge are called cations. Cations are formed when an atom loses its electron.
For example : [tex]\sf {Na}^{+} [/tex]
Anions - Atoms that carry negative charge are called anions. Anions are formed when an atom gains a electron.
For example : [tex]\sf {Cl}^{-} [/tex]
Question 8 of 21
Which nucleus completes the following equation?
The nucleus completing the following equation is option C: ₂₄⁵⁰Cr.
This reaction is a type of radioactive nuclei decay.
What is radioactive decay?Radioactive decay is the process by which unstable atomic nuclei undergo spontaneous transformations in order to achieve a more stable state. This is accomplished by the emission of particles and/or electromagnetic radiation from the nucleus. The decay may occur by several mechanisms, including alpha decay, beta decay, gamma decay, and electron capture.
In alpha decay, the nucleus emits an alpha particle, which consists of two protons and two neutrons, resulting in a daughter nucleus that has two fewer protons and two fewer neutrons than the original nucleus.
In beta decay, a neutron in the nucleus is converted into a proton and an electron, and the electron is then emitted from the nucleus as a beta particle. This results in the daughter nucleus having one more proton and one fewer neutron than the original nucleus.
In gamma decay, the nucleus emits a gamma ray, which is a high-energy electromagnetic radiation, without changing the number of protons or neutrons in the nucleus.
In electron capture, an electron from the inner shell of the atom is captured by the nucleus, and a proton in the nucleus is converted into a neutron. This results in the daughter nucleus having one fewer proton and one more neutron than the original nucleus.
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Joan wrote a science fiction story where the people only texted each other, and never talked. They still had vocal chords, but they could no
longer make sounds. Their vocal chords were
Answer:
Vestigial
Explanation:
The retention of genetically determined traits or structures that have partially or completely lost their ancestral purpose in a specific species is known as vestigiality. In most cases, evaluating the vestigality requires comparison with comparable traits in closely related species.
Write the complete equation for neutralization reactions for LiOh + HNO2
The complete equation for the neutralization reactions for the LiOH + HNO₂ is as :
LiOH + HNO₂ ----> LiNO₂ + H₂O
The Neutralization reaction is the reaction as in the chemical reaction in which the acid will reacts with the base and to produce the salt and the water molecule. The general equation of the chemical reaction is as :
HX + BOH --> BX + H₂O
The reaction with the LiOH and the HNO₂ is :
LiOH + HNO₂ ----> LiNO₂ + H₂O
There is the combination of the H⁺ ions and OH⁻ ions that will form the water.
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PLEASE ACTUALLY ANSWER THE WHOLE ASSIGNMENT FOR BRAINLIEST
The results of the lab activity showed that the larger the mass of the sun, the more likely at least one planet will fall into the habitable zone.
What effect does the mass of the Sun have on the orbits of Planets?The mass of the sun affects the orbits of planets in a solar system. When the mass of the sun is larger, the gravitational force between the sun and the planets is stronger, causing the planets to move at a slower pace around the sun.
Conversely, when the mass of the sun is smaller, the gravitational force is weaker, causing the planets to move at a faster pace.
Additionally, when Earth is closer to the sun, the gravitational force is stronger, causing its orbit to become faster, while a farther distance from the sun results in a slower orbit.
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Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.
Answer:
Explanation:
The chemical equation for the production of manganese from manganese (II) carbonate, oxygen, and aluminium can be represented as follows:
3MnCO3(s) + 3O2(g) + 4Al(s) → 3Mn(s) + 3CO2(g) + 2Al2O3(s)
In this equation, manganese (II) carbonate (MnCO3) reacts with oxygen (O2) and aluminium (Al) to produce manganese (Mn), carbon dioxide (CO2), and aluminium oxide (Al2O3). The equation is balanced with three molecules of manganese carbonate, three molecules of oxygen, and four molecules of aluminium reacting to produce three molecules of manganese, three molecules of carbon dioxide, and two molecules of aluminium oxide.
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