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tensa zangetsu [6.8K]
2 years ago
6

A. At STP, what is the volume of 708 mol of nitrogen gas? 708 mol = 708 mol X L B. A sample of hydrogen gas occupies 14.1 L at S

TP. Hov many moles of the gas are present? 14.1 L = 14.1 L X 11 mol​
Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
5 0

Answer:

A. 15859.2 L or 15900 L

B. 0.629 mol

Explanation:

At STP, one mole is equal to approximately 22.4 L

L or mL is volume, so you are attempting to solve for L or mL.

A.

708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)

B.

(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.

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Which element is chemically similar to chlorine?
ad-work [718]

bromine

Explanation:

halogens are a group of elemnts simlar to eachother

flourine, chlorine, and bromine

6 0
3 years ago
How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2
NNADVOKAT [17]

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂     →     CO₂

Number of moles of  CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

                          CO₂         :       O₂  

                              1           :         1

                            0.84       :       0.84

0.84 moles of oxygen are required.

6 0
3 years ago
Read 2 more answers
Tetrachloromethane,CC15,is classified as a
prohojiy [21]
Correct Answer: compound because the atoms of the elements are combined in a fixed proportion.
7 0
3 years ago
Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction
const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>

KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>

8 0
2 years ago
Question 1
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