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Triss [41]
3 years ago
11

A new engine is being designed that will carry 5.430 x 102 kilograms of methane, CH4, on board. How many liters of gaseous oxyge

n needs to be carried on board to ensure complete combustion? The oxygen will be stored at 3.500 atm of pressure and -170.84 °C. Use four significant figures in your answer. Hint: write out the balanced chemical reaction. Assume the gas behaves ideally.
(If you don't know don't answer. It takes up the points I worked really hard for for people who work equally as hard to help me understand these concepts)
Chemistry
1 answer:
algol [13]3 years ago
7 0

Answer:

yeet

Explanation:

You might be interested in
Calculate the ph of a 0.005 m solution of potassium oxide k2o
Alecsey [184]
First, we have to see how K2O behaves when it is dissolved in water:

K2O + H20 = 2 KOH

According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:

c(KOH) : c(K2O) = 1 : 2

So,

c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)

Now we can calculate pH:

pOH= -log c(OH⁻) = -log 0.01 = 2

pH= 14-2 = 12




3 0
3 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<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:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
PLEASE HELPPP ME
klasskru [66]
How many miles is in 2NA
6 0
3 years ago
Read 2 more answers
1. How many moles of oxygen are made if 12.0 moles of potassium chlorate react?
Alex787 [66]

Answer:

18

Explanation:

12*3=36

36/2=18

567890

3 0
3 years ago
A rock dropped in a graduated cylinder raises the level of water from 20 mL to 35 mL. The rock has a mass of 45 g. What is the d
Brut [27]
I think 15 I might be wrong
7 0
4 years ago
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