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3 years ago

Which of the following equations follows the law of conservation of mass?

Chemistry

2 answers:

Dovator [93]3 years ago

6 0

Answer:

option A = C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Explanation:

Law of conservation of mass:

This law stated that mass can not be created or destroyed in chemical reaction. It just changed from one to another form.

For example:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

28 g + 96 g = 88 g + 36 g

124 g = 124 g

This reaction correctly hold the law of conservation of mass.

Other options:

C + 4H₂ → CH₄

12 g + 8g = 16 g

20 g = 16 g

This reaction do not hold the law of conservation of mass.

3H₂O → 3H₂ + 3O₂

54 g = 6 g + 96 g

54 g = 102 g

This reaction do not hold the law of conservation of mass.

2Na + Cl → NaCl

46 g + 35.5 g = 58.5 g

81.5 g = 58.5 g

This reaction do not hold the law of conservation of mass.

valentina_108 [34]3 years ago

5 0

Answer:

Explanation:

look it up

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You have 1 mole of a gas at STP. If you apply the ideal gas law what is the approximate volume of the gas?

goldfiish [28.3K]

Answer:

A) 22.4L

Explanation:

we know, ideal gas law states

PV=nRT

V=nRT/P

At STP,

T= 273.15K P=1atm R=0.082L.atm/mol/K n=1 mole

V=(1*0.082*273.15)/ 1

V=22.4L

7 0

2 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0

3 years ago

Which of the following is an electrolyte?. A) potassium chloride. B) glucose. C) water D) mayonnaise.

kenny6666 [7]

Answer A )

Potassium chloride

K+ Cl-

hope this helps!.

8 0

3 years ago

Which is the limiting reactant if we start with 30.0 g Al and 30.0 h O2

lorasvet [3.4K]

Explanation:

The limiting reactant will be Al:

4Al + 3O₂ → 2Al₂O₃

The limiting reactant is the reactant in short supply in a chemical reaction.

Given parameters:

Mass of Al = 30g Molar mass = 27g/mol

Number of moles = $\frac{mass}{molar mass}$ = $\frac{30}{27}$

Number of moles of Al = 1.111 mole

Mass of O₂ = 30g, molar mass = 32g/mol

Number of moles = $\frac{30}{32}$ = 0.94mol

In the reaction:

4 moles of Al reacted with 3 moles of O₂

1.11moles of Al will require $\frac{1.11 x3}{4}$ = 0.83mole to react

But we have been given 0.94mole of O₂. This is more than required.

Therefore O₂ is in excess and Al is the limiting reactant.

Learn more:

Limiting reagents brainly.com/question/6078553

#learnwithBrainly

4 0

2 years ago

Based on the following reactants:

Doss [256]

(205) 872-9311 call me so I can help you out

5 0

3 years ago