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Mice21 [21]
3 years ago
9

Question 25 pts

Chemistry
2 answers:
Mama L [17]3 years ago
6 0
The answer is a. Hope this helped!
777dan777 [17]3 years ago
4 0
The answer is a, because the population number has increased.
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2.When copper (II) nitrate reacts with sodium hydroxide, copper (11) hydroxide is produced.
Ann [662]

Answer:

 

2Cu + S ~~~> Cu2S Copper (C) reacts with sulfur (S) to form copper sulfide as shown in the equation. A scientist adds 12.7 grams of Cu to 3.2 grams of S to start the reaction.

Explanation:

7 0
3 years ago
What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?
topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, R=0.08206\ L-atm/mol-K

Finding for V,

V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L

So, the volume of the gas is 34.55 L.

4 0
3 years ago
PLEASE HURRY AND TELL ME THE ANWERS
lesantik [10]

Answer:

Explanation:

3 0
3 years ago
Why do we use moles when calculating empirical chemical formulas, but not the mass?
Arada [10]

Answer:

because moles is the chemical unit*Avogadro's number, meaning it's the amount of atoms in the given unit.

Explanation:

8 0
3 years ago
Read 2 more answers
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
Alexeev081 [22]

Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

Thus, solving for the moles of the acid, we obtain:

n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

Regards.

7 0
3 years ago
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