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3 years ago

8

If one of the cars began with 5.6 liters of gasoline and oxygen, how much water and carbon dioxide do you think would be produce

d from that gasoline and oxygen?

1 answer:

lukranit [14]3 years ago

7 0

Answer

5.6 L

Explanation:

whatever you start with, you end with the same amount.

(LAW OF CONSERVATION)

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In which orbitals would the valence electrons for selenium (Se) be placed?

meriva

Answer:

Valence electrons of selenium will be placed in s and p-orbitals.

Explanation:

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3 years ago

Which type of change does the statement describe? Someone help fast

grandymaker [24]

Explanation:

bottom right = physical change

top left = chemical change

top right = physical change

bottom left = chemical change

<h3>A physical change is a change that goes from one form to another and physical changes can be reversed</h3><h3>example: water to ice or air to water</h3><h3 /><h3>A chemical change is when a substance combines with another substance and when it is combined it cannot be reversed</h3><h3>example: burning and or rusting</h3>

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3 years ago

Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio

boyakko [2]

Answer:

The concentration of CH₃OH in equilibrium is [CH₃OH] = <em>2,8x10⁻¹ M</em>

Explanation:

For the equilibrium:

CO (g) + 2H₂(g) ⇄ CH₃OH(g) keq= 14,5

Thus:

14,5 = $\frac{[CH_{3}OH]}{[CO][H_{2}]^2}$

In equilibrium, as [CO] is 0,15M and [H₂] is 0,36M:

14,5 = $\frac{[CH_{3}OH]}{[0,15][0,36]^2}$

Solving, the concentration of CH₃OH in equilibrium is:

<em>[CH₃OH] = 0,28M ≡ 2,8x10⁻¹ M</em>

I hope it helps!

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3 years ago

If 33.6 grams of KCl are dissolved in 192 grams of water, what is the concentration of the solution in percent by mass?21.2% KCl

Lelu [443]

The answer here would be 14.9 %KCl and here is how I can explain why:
33.5g / 225.6g x 100% = 14.9%
<span>you are looking for % of KCl in the solution, you have to add the mass of the KCl and water to get the total mass of the solution
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</span>

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4 years ago

Calculate the volume in milliliters of a iron(II) bromide solution that contains of iron(II) bromide . Round your answer to sign

miv72 [106K]

This is an incomplete question, here is a complete question.

Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.

Answer : The volume of iron(II) bromide solution is, $2.13\times 10^2mL$

Explanation : Given,

Concentration of iron(II) bromide = 1.29 mo/L

Moles of iron(II) bromide = 275 mmol = 0.275 mol

conversion used : 1 mmol = 0.001 mol

Now we have to calculate the volume of iron(II) bromide.

$\text{Volume of iron(II) bromide}=\frac{Moles of iron(II) bromide}}{\text{Concentration of iron(II) bromide}}$

Now put all the given values in this formula, we get:

$\text{Volume of iron(II) bromide}=\frac{0.275mol}{1.29mol/L}=0.213L=2.13\times 10^2mL$

Thus, the volume of iron(II) bromide solution is, $2.13\times 10^2mL$

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3 years ago