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blagie [28]
3 years ago
5

Balance the following equation

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
3 0
The answer would be D. Good luck my dude
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What is the empirical formula for a compound composed of 0.0683 mol of carbon ( C ), 0.0341 mol of hydrogen ( H ), and 0.1024 mo
liubo4ka [24]

The empirical formula for a compound composed of 0.0683 mol of carbon ( C ), 0.0341 mol of hydrogen ( H ), and 0.1024 mol of nitrogen ( N ) is  C_2HN_3.

<h3>What is the empirical formula?</h3>

An empirical formula tells us the relative ratios of different atoms in a compound.

Given data:

Moles of carbon = 0.0683 mol

Moles of hydrogen = 0.0341 mol

Moles of nitrogen = 0.1024 mol

Dividing each mole using the smallest number that is divided by 0.0341 moles.

We get:

Carbon= 2

Hydrogen=1

Nitrogen=3

The empirical formula for a compound isC_2HN_3.

Learn more about empirical formula here:

brainly.com/question/14044066

#SPJ1

4 0
2 years ago
Consider the equilibrium
Kamila [148]

Answer:

The answers are in the explanation

Explanation:

For the equilibrium:

B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq).

By LeChatelier's principle, the increase in the concentration of a reactant (for example) at equilibrium will change the system counteracting the increasing producing more product.

Thus,

A) Will the equilibrium constant for the reaction increase, decrease, or stay the same? Why? .

The equilibrium constant is a thermodynamic constant that stay the same with the addition of a compound.

B) Will the concentration of HB⁺(aq) increase, decrease, or stay the same? Why?

By LeChatelier's principle, the addition of B will induce the formation of more HB⁺(aq) increasing the concentration.

C) Will the pH of the solution increase, decrease, or stay the same? Why?

As the addition of B induce the increasing of OH⁻, the pH of the solution will increase.

I hope it helps!

7 0
3 years ago
What volume of a 0.124 M KOH solution neutralizes 23.4 mL of 0.206 M HCl solution?
alexgriva [62]
A) 15.9 mL I hope it is if not I’m sorry
6 0
3 years ago
Yo sum help would be sick
Agata [3.3K]

Answer:

A

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the mass of MgCO3 (84.31 g/mol) precipitated by mixing 10.0 mL of a 0.300 M Na2CO3 solution with 6.00 mL of 0.0400 M M
I am Lyosha [343]

Answer:

m_{MgCO_3}=0.0202molMgCO_3

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3

In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3}  \\\\m_{MgCO_3}=0.0202molMgCO_3

Regards.

7 0
3 years ago
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