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lisov135 [29]
3 years ago
5

137 lbs = how much ounces

Chemistry
2 answers:
vivado [14]3 years ago
6 0
137 lbs = 2192 oz , hope this helps :)
ahrayia [7]3 years ago
6 0

Answer:

2192 ounces

Explanation:

In order to find the answer to this question you need to remember that they're 16 ounces in every pound so multiplying 16 ounces by 137 will give you your answer.

137 lbs

1 lb = 16

16\times137 = 2192

= 2192 ounces

Hope this helps.

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The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces
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Answer:

1. NaN₃(s) → Na(s) + 1.5 N₂(g)

2. 79.3g

Explanation:

<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>

NaN₃(s) → Na(s) + 1.5 N₂(g)

<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>

First, we have to calculate the moles of N₂ from the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm.(43.0L)}{(0.08206atm.L/mol.K).286.2K} =1.83mol

The moles of NaN₃ are:

1.83molN_{2}.\frac{1molNaN_{3}}{1.5molN_{2}} =1.22molNaN_{3}

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

1.22mol.\frac{65.01g}{mol} =79.3g

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
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<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

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