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alisha [4.7K]
3 years ago
9

A metal sample weighs 57.3g and has volume of 6.3mL what is the density of the sample

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

<h2>9.10 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{57.3}{6.3}  \\  = 9.095238...

We have the final answer as

<h2>9.10 g/mL</h2>

Hope this helps you

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Answer:

B.) 1.3 atm

Explanation:

To find the new pressure, you need to use Gay-Lussac's Law:

P₁ / T₁ = P₂ / T₂

In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.

P₁ = 1.2 atm                                    P₂ = ? atm

T₁ = 20 °C + 273 = 293 K              T₂ = 35 °C + 273 = 308 K

P₁ / T₁ = P₂ / T₂                                             <----- Gay-Lussac's Law

(1.2 atm) / (293 K) = P₂ / (308 K)                  <----- Insert values

0.0041 = P₂ / (308 K)                                   <----- Simplify left side

1.3 = P₂                                                         <----- Multiply both sides by 308

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A buffer can be prepared by mixing two solutions. Determine if each of the following mixtures will result in a buffer solution o
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Answer:

The answers are in the explanation

Explanation:

A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:

<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.

<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.

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<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.

<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).

I hope it helps!

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3 years ago
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