Answer:
When you fill in 2 reactants, you will have 6 N atoms but on the side of the product there are only 2 N atoms, so the equation cannot be balanced and for the equation to be equal, you need to fill 3 into the part N2 to have 6 N atoms
Explanation:
The given reaction unbalanced because of the number of nitrogen is unequal in both side. Three molecules of N₂ should be there in the product side of this decomposition reaction.Thus option C is correct.
What is decomposition reaction?
A decomposition reaction is the one in which a compound decomposes into its constituent compounds or molecules.It is mostly exothermic thus, release heat energy.
In balanced reaction, each elements have to be equal in number in both reactant and product side. The imbalanced reaction can be balanced multiplying with simple intergers.
In the given reaction, sodium is balanced and nitrogen is not. There are 6 nitrogen in the reactant side and only 2 in the product side. Hence, the product side have to multiplied by 3. Hence, Three molecules of N₂ should be produced during the decomposition. Thus, option C is correct.
To find more on decomposition reactions, refer here:
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Your question is incomplete.But your complete question is as follows:
An unbalanced chemical equation is shown:
2NaN3 → 2Na + N2
Which of the following statements explains why the equation is not balanced?
A. Four molecules of N2 should be produced during the synthesis reaction.
B. Four molecules of N2 should be produced during the decomposition.
C. Three molecules of N2 should be produced during the decomposition.
D. Three molecules of N2 should be produced during the synthesis reaction.
C. Three molecules of N2 should be produced during the decomposition.
Molecules have Question 8 options: A) only kinetic energy. B) neither kinetic nor potential energy. C) only potential energy. D) both potential and kinetic energy.
What is the freezing point in °C) of a 0.195 m
aqueous solution of K2S?
Enter your rounded answer with
3 decimal places.
K for water = 1.86 °C/m
Answer:
[tex]T_S=-1.09\°C[/tex]
Explanation:
Hello!
In this case, since the freezing point depression for a solution is computed via:
[tex](T_S-T_W)=-imK_f[/tex]
Whereas TW is the freezing temperature of water, TS that of the solution, i the van't Hoff's factor (3 for K2S as it ionizes properly), m the molality of the solution and Kf the freezing point constant of water. Thus, we plug in to obtain:
[tex](T_S-0\°C)=-3*0.195m*1.86\frac{\°C}{m}\\\\T_S=-1.09\°C[/tex]
Best regards!
The correct answer is -1.088. (Don't forget the negative sign).
What mass of nitrogen occupies a volume of 11.2 L, if 4.20 g of nitrogen occupies 100. L?
Answer:
0.47 g
Explanation:
From the question given above, we obtained the following information:
4.20 g of nitrogen occupies 100 L.
We can obtain the mass of nitrogen that occupied 11.2 L by doing the following:
4.20 g of nitrogen occupies 100 L.
Therefore, Xg of nitrogen will occupy 11.2 L i.e
Xg of nitrogen = (11.2 × 4.2) / 100
Xg of nitrogen = 0.47 g
From the calculation made above, 0.47 g of nitrogen will occupy 11.2 L
What happened to the celery if you put in food color water
Answer:
the celery would change colour!
Explanation:
Which one is a decomposition reaction?
Answer:
b no
Explanation:
because it is decomposing into two elements
What is the degree of oxidation of a simple substance
Answer: The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl– has an oxidation state of -1. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2.
Hope this helps...... Stay safe and have a Merry Christmas!!!!!!!!!!! :D
Explanation:
Suppose we have 100 g of each of the following substances. Which sample contain the greatest number of moles (F.W. = Formula Weight) A. HCI, F.W. = 36.5 B. H_2O, F.W. = 18.0 C. MgCO_3 F.W. = 84.3 D. AlCI_3, F.W. = 133.3 E. NaCl, F.W. = 58.4
Answer:
100 g of water has the highest number of moles
Explanation:
Recall that the number of moles is obtained as given mass/formula weight
For HCl;
number of moles = 100g/36.5g/mol = 2.7 moles
For H2O;
number of moles = 100g/18g/mol = 5.5 moles
For MgCO3
number of moles = 100g/84.3 g/mol = 1.2 moles
For AlCl3
number of moles = 100g/133.3g/mol = 0.75 moles
For NaCl
number of moles = 100g/58.4 g/mol = 1.7 moles
In which scenarios would the amount of substance remaining be 200 mg? Select all that apply. 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years 1,000 mg of a radioactive substance with a half-life of 5 years after decaying for 20 years 300 mg of a radioactive substance with a half-life of 1 year after decaying for 1 year 600 mg of a radioactive substance with a half-life of 6 hours after decaying for 18 hours 400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours
Answer:
e) 400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours
a) 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years.
Explanation:
a) 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years.
Number of half lives passed = Time elapsed/ half life
Number of half lives passed = 4 year/2 year
Number of half lives passed = 2
at time zero = 800 mg
At first half life = 800 mg/2 = 400 mg
At 2nd half life = 400 mg/2 = 200 mg
b) 1,000 mg of a radioactive substance with a half-life of 5 years after decaying for 20 years
Number of half lives passed = Time elapsed/ half life
Number of half lives passed = 20 year/ 5 year
Number of half lives passed = 4
at time zero = 1000 mg
At first half life = 1000 mg/2 = 500 mg
At 2nd half life = 500 mg/2 = 250 mg
At 3rd half life = 250 mg/2 = 125 mg
At 4th half life = 125 mg/2 = 62.5 mg
c) 300 mg of a radioactive substance with a half-life of 1 year after decaying for 1 year
Number of half lives passed = Time elapsed/ half life
Number of half lives passed = 1 year/ 1 year
Number of half lives passed = 1
at time zero = 300 mg
At first half life = 300 mg/2 = 150 mg
d) 600 mg of a radioactive substance with a half-life of 6 hours after decaying for 18 hours
Number of half lives passed = Time elapsed/ half life
Number of half lives passed = 18 hours / 6 hours
Number of half lives passed = 3
at time zero = 600 mg
At first half life = 600 mg/2 = 300mg
At 2nd half life = 300 mg/2 = 150 mg
At 3rd half life = 150 mg/2 = 75 mg
e) 400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours
Number of half lives passed = Time elapsed/ half life
Number of half lives passed = 8 hours / 8 hours
Number of half lives passed = 1
at time zero = 400 mg
At first half life = 400 mg/2 = 200 mg
In which type of mirror do you see the most realistic virtual image
Plane
Convex
Concave
Duplex
Phan
convex
concave
duplex
What type of radioactive decay will the isotopes 13B and 188Au most likely undergo?
a. Beta emission, positron emission
b. Beta emission, beta emission
c. Positron emission, beta emission
d. Positron emission, positron emission
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
Radioactive decay refers to the loss of energy by an unstable atomic nucleus as radiation. The substance having unstable nuclei is radioactive.
The correct answer is:
Option B. Beta emission, beta emission
The radioactive decay of 13B and 188 Au can be explained as:
1. Neutron proton ratio refers to the particular nuclei that will undergo what type of radioactive decay.
For B-13, there are 8 neutrons and 5 protons, thus, the ratio is 1.6.For Au -188, there are 109 neutrons and 79 protons, thus, the ratio is 1.4.2. In the B-13 element, the N/P ratio is beyond the stability of the atom, thus, it will emit beta emission to reduce the N/P ratio.
3. Similarly, the N/P ratio of the Au-188 is above the stability of the atom, thus, it undergoes the beta emission to reduce the N/P ratio.
Thus, the correct answer is Option B.
To know more about radioactive decay, refer to the following link:
https://brainly.com/question/14077220
Does the octet rule state that all elements except Hydrogen and Helium want eight (8) valence electrons?
Answer:
yes
Explanation:
What is the final concentration of a hydrogen peroxide solution if 600 mL of a 1.2M solution was diluted to 1.0 L?
Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.
Solution:
M1V1 = M2V2
(1.6 mol/L) (175 mL) = (x) (1000 mL)
x = 0.28 M
Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.
Problem #2: You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?
Solution:
M1V1 = M2V2
(x) (2.5 L) = (1.2 mol/L) (10.0 L)
x = 4.8 M
Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.
Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?
Solution:
M1V1 = M2V2
(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)
5x = 48 + 0.3x
4.7x = 48
x = 10. mL (to two sig figs)
The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.
Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?
Solution:
Here is the first way to solve this problem:
M1V1 + M2V2 = M3V3
(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)
Where x is volume of 5.65 M HCl that is added
(0.250 + x) is total resultant volume
0.8875 + 5.65x = 1.125 + 4.50 x
1.15x = 0.2375
x= 0.2065 L
Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L
Total amount = 456.5 mL
Here is the second way to solve this problem:
Since the amount of 5.65 M added is not asked for, there is no need to solve for it.
M1V1 + M2V2 = M3V3
(3.55) (250) + (5.65) (x − 250) = (4.50) (x)
That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.
Problem #5: A 40.0 mL volume of 1.80 M Fe(NO3)3 is mixed with 21.5 mL of 0.808M Fe(NO3)3 solution. Calculate the molar concentration of the final solution.
A baseball struck by a bat flies away from the batter because
of:
a) Normal force
b) Friction
c) Electrical Force
O d) Applied Force
Answer:
the answer is applied force
What is the characteristic of atoms
Answer:
Most of the atom is empty space. The rest consists of a positively charged nucleus of protons and neutrons surrounded by a cloud of negatively charged electrons. The nucleus is small and dense compared with the electrons, which are the lightest charged particles in nature.
Which of the following equations is balanced?
Answer:
wheres the image
Explanation:
The base that will NOT combine with 2-deoxyribose to form a nucleic acid is _____.
Answer:
Uracil
Explanation:
The base that will NOT combine with 2-deoxyribose to form a nucleic acid is Uracil.
2-deoxyribose is a pentose sugar found in the DNA (Deoxyribonucleic acid). It is devoid of oxygen in its 2' position. The bases found in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine, Guanine, and Cytosine are also found in RNA (Ribonucleic acid). Thymine is not present in RNA, it is only found in DNA. The base found in RNA is Uracil which in turn is not present in DNA. The five carbon sugar present in RNA is ribose which combines with Uracil.
Q. The entropy of the system will usually increase when
answer choices
A)a molecule is broken into two or more smaller
molecules
B)a reaction occurs that results in an increase in
the number of moles of gas
C)a solid changes to a liquid
D)all of these
Answer:
D
Explanation:
The enthalpy change of formation of carbon dioxide is –394 kJ mol–1. The enthalpy change of formation of water is –286 kJ mol–1. The enthalpy change of formation of methane is –74 kJ mol–1. What is the enthalpy change of combustion of methane?
The enthalpy change of combustion of methane : -892 kJ/mol
Further explanationThe change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction for combustion of Methane (CH₄) :
CH₄ + 2O₂ → CO₂ + 2H₂O.
The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero(∆Hf °O₂ =0)
∆Hf ° CO₂ = -394 kj/mol
∆Hf ° H₂O=-286 kj/mol
∆Hf ° CH₄=-74 kJ/mol
The enthalpy change of combustion of methane :
[tex]\tt \Delta H_{rxn}=\Delta H_f~CO_2+2.\Delta H_fH_2O-(\Delta H_fCH_4)\\\\\Delta H_{rxn}=(-394-2.-286)-(-74)\\\\\Delta H_{rxn}=-892~kJ/mol[/tex]
Answer:
-655
Explanation:
ye
How many grams are 1.2 x 10^-2 moles of K3PO4
Answer:
2.5471951319999997
Explanation:
Hybridization of carbons in CH3-CH= C = CH2 is
a) sp3sp3, sp2 sp2
b) sp3. sp2, sp, sp3
c) sp3, sp3, sp2, sp3
d) sp3, sp2, sp, sp1
Answer:
A carbon atom bound to three atoms (two single bonds, one double bond) is sp2 hybridized and forms a flat trigonal or triangular arrangement with 120° angles between bonds. Notice that acetic acid contains one sp2 carbon atom and one sp3 carbon atom.
3. A reaction of the type X → Y + Z has a rate constant k = 3.6 x 10-5 M/s. Determine
molarity of x after a reaction time of 30.0 min if the initial concentration of x is
0.096M
(3 marks)
Answer:
0.16 M.
Explanation:
Hello!
In this case, since the reaction can be determined as zeroth-order because of the units of M/s, we can write the rate law as:
[tex][X]=[X]_0-kt[/tex]
In such a way, since we need the initial concentration of X, we proceed as follows:
[tex][X]_0=[X]+kt[/tex]
So we plug in to obtain:
[tex][X]_0=0.096M+3.6x10^{-5}\frac{M}{s}*30.min*\frac{60s}{1min}[/tex]
[tex][X]_0=0.16M[/tex]
Best regards!
How many neutrons are there in a neutral argon atom? Show how you found your answer on your scratch paper.
What happens when sodium and sulfur combine?
A) each sodium atom gains one electron
B) each sulfur atom loses one electron
C) each sodium atom loses one electron
D) each sulfur atom gains one electron
Answer:
BBBBBBBB ITS B
Explanation:
Molecules Have?
A. Only kinetic energy
B. Only potential energy
C. Both kinetic and potential energy
D. Neither potential nor kinetic
Answer:
C
Explanation:
Because molecules of solids don't move
while molecules of liquids and gases move.
I Hope this helps!!.
A fish company delivers 15 kg of salmon 4.7 kg Of crab and 1.76 kg of oysters to your seafood restaurant what is the total mass in kilograms of the seafood
Answer:
21.46 kg.
Explanation:
Hello!
In this case, when a variety of measurements are said to contribute to a total, we understand we need to add them all up in order to get that total; in such a way, since the given seafood is composed by 15 kg of salmon, 4.7 kg of crab and 1.76 kg of oysters, the total mass turns out:
[tex]m_T=15kg+4.7kg+1.76kg\\\\m_T=21.46kg[/tex]
Best regards!
How many molecules of H2 will react with one molecule of the following triglyceride? CH2-O-C-(CH)14CH CH O-C-(CH)CH CH (CH)-CH CH2-O-C-(CH)16CH a. 1 b.3 c. 4 d. 6
Answer:
1
Explanation:
The full question is shown in the image attached.
If we look at the structure closely, we will discover that there is only one C=C double bond in the molecule.
It is worthy of note that hydrogen (H2) only reduces the C=C double bond. it does not affect any other bond in the molecule.
Hence, only one molecule of H2 is required to reduce one C=C bond present.
The equilibrium constant, Kc, for the following reaction is 1.29E-2 at 600 K. COCl2(g) CO(g) Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl2]
Answer:
At Equilibrium
[COCl₂] = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Explanation:
Given that;
equilibrium constant Kc = 1.29 × 10⁻² at 600k
the equilibrium concentrations of reactant and products = ?
when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]
Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M
COCl₂(g) ----------> CO(g) + Cl₂(g)
0.280 0 0 ------------ Initial
-x x x
(0.280 - x) x x ----------- equilibrium
we know that; solid does not take part in equilibrium constant expression
so
KC = [CO][Cl₂] / COCl₂
we substitute
1.29 × 10⁻² = x² / (0.280 - x)
0.0129 (0.280 - x) = x²
x² = 0.003612 - 0.0129x
x² + 0.0129x - 0.003612 = 0
x = -b±√(b² - 4ac) / 2a
we substitute
x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]
x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2
x = [-0.0129 ± 0.1209] / 2
Acceptable value of x =[ -0.0129 + 0.1209] / 2
x = 0.108 / 2
x = 0.054
At equilibrium
[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Which segment of this graph shows a decreasing velocity?
Answer:
The segment of this graph that shows a decreasing velocity is the EF segment. AB and DE segments show that whatever was moving, in these segments was stopped. The segment BC shows an increase in velocity - the metres travelled were always more as the time went by.
Explanation:
Which actions can be taken to plan for a drought check all that apply
Answer:
Can you attach a picture?
Explanation:
Oxidation of a secondary alcohol results in the formation of a(n):________. a) tertiary alcohol b) acid c) aldehyde d) ketone
Answer:
A. Tertiary alcohol
Explanation:
Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute Sulphuric acid, you get propanone formed. Playing around with the reaction conditions makes no difference whatsoever to the product.