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3 years ago

If you have 5 grams of carbon and 6 grams of hydrogen and you want to make methane, what is reaction limiting?

1 answer:

5 0

Methane's formula is CH4 I don't get the question but if those are the quantities of 1 of those metals, that means that you don't have enough grams of carbon. You will need 5g x 4 = 20grams of carbon.

Hope that's what you were looking for.

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Need help on B and C please!!

puteri [66]

B. The reason the temperature experienced no change in group c is because it was likely the control group.
I cannot read question c, the monitor refresh is obscuring the text.

5 0

3 years ago

I need help on thiss

Sergio039 [100]

The reaction is not balanced

<h3>Further explanation</h3>

Given

Reaction

2Fe(s)+3O₂(g)⇒2Fe₂O₃(s)

Required

The number of atoms

Solution

In a balanced chemical equation, the number of atoms in the compound that reacts (the reactants and products) will have the same number

Reactants : Fe(s)+O₂(g)

Fe = 2 atoms

O = 3 x 2 = 6 atoms

Products : Fe₂O₃(s)

Fe = 2 x 2 = 4 atoms

O = 2 x 3 = 6 atoms

The reaction is not balanced because the number of Fe atoms is not the same

The balanced reaction should be:

4Fe(s)+3O₂(g)⇒2Fe₂O₃(s)

8 0

3 years ago

How many grams are in 2.3 moles of N?

jok3333 [9.3K]

The amount of grams that are in 2.3 moles of N = 32.223 or 32/100
Because there are 14.01 grams per mile of nitrogen atoms.
So…
14.01 x 2.3= 32.223
Hope this helps :)

6 0

3 years ago

Read 2 more answers

What is the name of this compound

lys-0071 [83]

Answer:

horizontal

Explanation:

5 0

3 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0

3 years ago