Changing capacity science definition

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

3 years ago

On the first day of your new job as a chemist, you are given a bottle of magnesium sulfate and asked to make 30 mL of 0.3 M MgSO

Lady bird [3.3K]

Answer:

The answer to your question is 15500 mg of MgSO₄ 7H₂O

Explanation:

Data

Volume = 30 ml

Concentration = 0.3 M

Formula MgSO4∗7H2O

Process

1.- Calculate the number of moles needed

Molarity = moles / volume

Solve for moles

moles = Molarity x volume

Substitution

moles = 0.3 x 0.03 l

moles = 0.009 moles of MgSO₄

2.- Calculate the molecular mass of MgSO4∗7H2O

molecular mass = 24 + 32 + 64 + 14 + 112

= 246 g

3.- Get the proportion MgSO4 :7H2O, this proportion is 1 : 7

4.- Calculate the amount of MgSO4∗7H2O

$0.009 mol of MgSO4 \frac{7 MgSO4 7H2O}{1 mol MgSO4} \frac{246 g MgSO4 7H2O}{1 mol MgSO4 7H2O}$

Simplification

15.5 g = 15500 mg

5 0

3 years ago

What are differences and similarities between water and atoms?Explain

REY [17]

Answer:

Atom may be defined as the smallest particle of an element which does not exist free in nature but takes part directly in chemical combinations.

Atom of any element is represented by the symbol of that element. For example, hydrogen atom is represented by H, sodium atom is represented by Na, etc.

Molecule may be defined as the smallest particle of an element or compound which exists free in nature but does not participate directly in chemical combinations.

Molecule of any substance is represented by their molecular formula. For example, water molecule is represented by H2O. It tells that each water molecule contains two atoms of hydrogen and one atom of oxygen.

8 0

3 years ago

O Previous

Ede4ka [16]

Answer:

what is scientific model

Explanation:

4 0

3 years ago

In your own words, please describe the difference between a complete ionic equation and a net ionic equation.

beks73 [17]

Explanation:

In the molecular equation for a reaction, all of the reactants and products are represented as neutral molecules (even soluble ionic compounds and strong acids). In the complete ionic equation, soluble ionic compounds and strong acids are rewritten as dissociated ions.

The net ionic equation is a chemical equation for a reaction that lists only those species participating in the reaction. The net ionic equation is commonly used in acid-base neutralization reactions, double displacement reactions, and redox reactions.

6 0

3 years ago