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3 years ago

What is deposition ?

Chemistry

2 answers:

Angelina_Jolie [31]3 years ago

8 0

Answer:

the act of deposing someone especially a monarch

slavikrds [6]3 years ago

7 0

The act of depositing material, especially by a natural process; the resultant deposit.

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Can someone please help me with this question also explain the answers I am so confused thank you.

Archy [21]

The theoretical yield of H₂S is 13.5 g.

The percent yield is 75.5 %.

<h3>What is the theoretical yield of H₂S from the reaction?</h3>

The equation of the reaction is given below:

FeS + 2 HCl → FeCl₂+ H₂S

Moles of FeS reacting = mass/molar mass

Molar mass of FeS = 88 g/mol

Moles of FeS reacting = 35/88 = 0.398 moles

Moles of H₂S produced = 0.398 moles

Molar mass of H₂S = 34 g/mol

Mass of H₂S produced = 0.398 * 34 = 13.5 g

Theoretical yield of H₂S is 13.5 g.

Percent yield = actual yield/theoretical yield * 100%

Actual yield of H₂S = 10.2 g

Percent yield = 10.2/13.5 * 100%

Percent yield = 75.5 %

In conclusion, the actual yield is less than the theoretical yield.

Learn more about percent yield at: brainly.com/question/8638404

#SPJ1

4 0

1 year ago

Use the following equation to answer the questions below:

Gala2k [10]

Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass * Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

3 0

3 years ago

How many moles of atoms are in 7.00 g of 13c? express your answer numerically in moles?

Nadusha1986 [10]

Molar mass of 13c = 13 grams
number of moles = mass / molar mass
therefore,
number of moles = 7 / 13
To know the number of atoms in 7/13 moles, we simply multiply the number of moles by Avogadro's number as follows:
number of atoms = (7/13) x 6.022 x 10^23 = 3.2426 x 10^23 atoms

3 0

3 years ago

Read 2 more answers

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation

nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ$ ..[2]

$H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}$

$\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_{4}=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

5 0

2 years ago

Compare and contrast an electric generator and a battery??

anygoal [31]

Both generators and batteries both convert a form of energy into electrical energy. In a battery, a chemical reaction takes place which converts chemical energy into electrical energy. In a generator however, many times mechanical energy is being converted into electrical energy. A process called electromagnetic induction can take place in some generator which is where an electromagnet is used to help conduct electricity. hope this helped!!!

3 0

3 years ago