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Assoli18 [71]
3 years ago
10

A gas in a balloon at constant pressure has a volume of 185 mL at -125*C. What is its volume at 31.0*C? Show all work including

what equation or gas law you use.
Chemistry
1 answer:
miv72 [106K]3 years ago
4 0

Answer:

380 mL is the new volume

Explanation:

At constant pressure.

V₁ / T₁ = V₂ / T₂

Temperature must be in Absolute Values (T°K = T°C + 273)

-125°C + 273 = 148 K

31°C + 273 = 304 K

185 mL / 148 K = V₂ / 304 K

V₂ = (185 mL / 148 K) . 304 K → 380 mL

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3 years ago
4 Na + O2 → 2 Na2O<br><br> 6.79 moles of O2 will react to form how many moles of Na2O?
Nataly_w [17]

Answer:

13.94moles of Na₂O

Explanation:

The balanced reaction expression is given as:

        4Na  +  O₂  →   2Na₂O

Given parameters:

Number of moles of O₂ = 6.97moles

Unknown:

Number of moles of Na₂O

Solution:

 To solve this problem;

            1 mole of O₂  will produce 2 moles of Na₂O ;

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A student uses a pH meter to measure the acidity of a water sample from a lake. For what purpose is the student MOST likely test
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Answer : The correct option is, C.

Explanation :

pH meter : It is an electric device which measure the concentration of hydrogen-ion in the solution. It also measure the acidity or alkalinity of water in the solution.

Therefore, the purpose of student for testing the lake water is to understand how people use it in their daily-life. As this will help him to analyse the quality of lake water whether, it is fit for use or not.

4 0
3 years ago
A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4
natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

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The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

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The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M

8 0
3 years ago
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