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makvit [3.9K]
3 years ago
6

How many protons, neutrons, and electrons are there in one atom of 'H ?

Chemistry
1 answer:
kolezko [41]3 years ago
8 0

Answer:

1

Explanation:

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What is the percent yield of NaCl if 31g of CuCl2 reacts with excess NaNo3 to produce 21.2g of NaCl
Valentin [98]
The answer is 78.7% yield.
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3 years ago
Identifying Qualitative and quantitative Data
lyudmila [28]

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qualitative data :           quantitative data :      

circular in shape            75 colonies ...          

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Explanation:

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7 0
3 years ago
A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the
Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}

P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

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4 0
2 years ago
What volume will 2.5 mol of a gas at STP occupy ?
kvv77 [185]

Answer:

56

Explanation:

1 mole of gas at STP occupies 22.4 L of the gas

2.5 mole of the gas at STP occupies 22.4×2.5 L of the gas

so 2.5 mole of the gas at STP occupies 56 L of the gas .

6 0
2 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
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