Ask question

Ask question

All categories

3 years ago

11

In the following reaction, when a surplus of hydrazine (N2H4) reacts with 20 liters of hydrogen peroxide (H2O2), how many liters

of nitrogen gas will be released, given that both gases are at STP?
N2H4 + 2H2O2 →N2 + 4H2O

10 liters

0.1 liters

40 liters

20 liters

2 answers:

Ghella [55]3 years ago

5 0

The answer is 10 liters.

Kaylis [27]3 years ago

5 0

The correct answer is

10

:)

You might be interested in

Moon rocks collected on the moon were returned to earth are made of basalt, a light volcanie rock, have a density of 2.7 g/cm?.

Marina86 [1]

Answer:

160cm³

Explanation:

$V = \frac{m}{D} = \frac{432}{2.7} = 160$

5 0

3 years ago

The molar mass of SO3 is grams. in significant figures

Vikentia [17]

Answer:

80.066 g/mol

Explanation:

I looked it up

6 0

3 years ago

Read 2 more answers

If a 66.5 g sample of ammonium nitrate (NH4NO3) is dissolved in enough water to make 315 mL of solution, what will be the molari

11111nata11111 [884]

Answer:

2.64 M

Explanation:

To find the molarity, you need to (1) convert grams to moles (via molar mass), then (2) convert mL to L, and then (3) calculate the molarity (via molarity ratio). The final answer should have 3 sig figs to match the sigs figs of the given values.

(Step 1)

Molar Mass (NH₄NO₃): 2(14.007 g/mol) + 4(1.008 g/mol) + 3(15.998 g/mol)

Molar Mass (NH₄NO₃): 80.04 g/mol

66.5 grams NH₄NO₃ 1 mole
--------------------------------- x ---------------------- = 0.831 moles NH₄NO₃
80.04 grams

(Step 2)

1,000 mL = 1 L

315 mL 1 L
-------------- x ------------------ = 0.315 L
1,000 mL

(Step 3)

Molarity = moles / volume

Molarity = 0.831 moles / 0.315 L

Molarity = 2.64 M

7 0

1 year ago

Breathing equipment used by rescue workers needs to capture the CO2 the humans breath out and produce O2 for them to breath in,

const2013 [10]

Answer: $4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for reaction of potassium superoxide with carbon dioxide to produce oxygen and potassium carbonate will be:

$4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

8 0

2 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago