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2 years ago

Which contains more molecules: a mole of hydrogen or a mole of chlorine

Chemistry

2 answers:

Snowcat [4.5K]2 years ago

7 0

Answer:

Neither.

Step-by-step explanation:

A mole of particles contains <em>Avogadro's number</em> of particles. Thus,

A mole of hydrogen contains 6.023 × 10²³ hydrogen molecules.

A mole of chlorine contains 6.023 × 10²³ chlorine molecules.

A mole of hydrogen and a mole of chlorine contain the same number of molecules.

Maurinko [17]2 years ago

3 0

b is answer good luck.

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A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

Grace [21]

Answer:

0.21 g

Explanation:

The equation of the reaction is;

NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)

Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles

Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles

Since the reaction is 1:1, NaCl is the limiting reactant.

1 mole of NaCl yields 1 mole of AgCl

0.00147 moles of NaCl yields 0.00147 moles of AgCl

Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol

= 0.21 g

3 0

2 years ago

The volume of a gas decreases to half of its original volume, but the gas maintains the same number of moles and temperature. Ac

Leya [2.2K]

It will double

Key
P = pressure of the gas
n= number of moles of the gas
V= volume of the gas
R = the gas constant
T= absolute temperature of the gas
If V is halved when n and T remain the same

4 0

2 years ago

What are the 3 types of evidence of Pangaea? *

seraphim [82]

Explanation:

fit of the continents, paleoclimate indicators, truncated?

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2 years ago

What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.4

sesenic [268]

Answer: The concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the $CaCl_2$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $AlCl_3$

We are given:

$n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL$

Putting all the values in above equation, we get

$M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M$

The concentration of $Cl^-$ ions in the resulting solution will be same as the molarity of solution which is 1.16 M.

Hence, the concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

6 0

2 years ago

Use the image above and describe the

Lina20 [59]

Answer:

If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.

Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.

4 0

2 years ago