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seraphim [82]
3 years ago
12

Before solving for moles of hydrogen produced you must first solve for moles of zinc how do you convert grams of zinc to moles o

f zinc?
Chemistry
2 answers:
Alika [10]3 years ago
5 0

Converting grams to moles of zinc,

; Use the formula...,

moles = (given mass in grams)/relative atomic mass of zinc

geniusboy [140]3 years ago
3 0

Answer: Multiply 10 grams of Zn x (1 mol Zn/65.39 g)  

I hope this helps!

You might be interested in
The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
Elden [556K]

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

6 0
3 years ago
2. What pressure is required to compress 196.0 L of air at 1.00 atmosphere into a cylinder
damaskus [11]

Answer:

<h2>7.54 atm </h2>

Explanation:

The required pressure can be found by using the formula for Boyle's law which is

P_2 =  \frac{P_1V_1}{V_2}  \\

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

From the question we have

P_2 =  \frac{1 \times 196}{26}  =  \frac{196}{26}   \\ = 7.538461...

We have the final answer as

<h3>7.54 atm </h3>

Hope this helps you

3 0
2 years ago
Calculate each of the following quantities.<br> (a) mass in kilograms of 3.7 x 1020 molecules of NO2
HACTEHA [7]

Answer:

The answer to your question is:  0.028 kg of NO2

Explanation:

Data

3.7 x 10²⁰ molecules of NO2 in kg

MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg

                   1 mol of NO2 ---------------------  6.023 x 10 ²³ molecules

                   x                     --------------------- 3.7 x 10²⁰ molecules

                   x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

                   x = 0.00061 mol

         

                 1 mol of NO2 ---------------------  46 kg of NO2

                 0.00061 mol     ------------------    x

                 x = 0.00061 x 46/1

                x = 0.028 kg of NO2

7 0
3 years ago
In this reaction, what is the correct coefficient for hydrogen gas? ? h2 + ? o2 ? ? h2o
iragen [17]
To balance a chemical reaction, it is important to remember that the number of atoms of each element in the reactants and the product side should be equal. This is to follows the law of conservation of mass where mass cannot be created or destroyed. So, the total mass that is used to react should have the same value of the total mass of the substances produced from the reactants. The balanced chemical reaction would be written as follows:

<span> 2h2 + o2 = 2h2o

Reactant     =    Product
H = 4           =          4 
O = 2          =           2

Therefore, the correct coefficient for the hydrogen gas would be 2.</span>
6 0
2 years ago
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
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