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2 years ago

How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4

Chemistry

1 answer:

blsea [12.9K]2 years ago

3 0

Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

C₁V₁=C₂V₂

Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

C₁ = 0.25 M

V₁ = 100 L

C₂ = 7.5 M

V₂ = ?

We input the data:

0.25 M * 100 L = 7.5 M * V₂

V₂ = 3.33 L

Thus the answer is 3.33 liters.

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What’s the Difference between attractive forces and ionic/covalent/metallic bonds

Deffense [45]

Answer:

Explanation:The main difference between ionic covalent and metallic bonds is their formation; ionic bonds form when one atom provides electrons to another atom whereas covalent bonds form when two atom shares their valence electrons and metallic bonds form when a variable number of atoms share a variable number of electrons in a metal lattice.

7 0

2 years ago

A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t

Serga [27]

Explanation:

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

$1.00 M=\frac{m}{40 g/mol\times 1.00 L}$

$m=1.00 M\times 40 g/mol\times 1.00 L = 40. g$

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

$M_1V_1=M_2V_2$ (dilution equation)

Molarity of the NaOH solution = $M_1=2.00 M$

Volume of the solution = $V_1=?$

Molarity of the NaOH solution after dilution = $M_2=1.00 M$

Volume of NaOH solution after dilution= $V_2=1 L$

$M_1V_1=M_2V_2$

$V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L$

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

8 0

3 years ago

How many moles of ammonia can be formed from 4.0 mol H2

Bess [88]

Answer: 1. 2.7 moles of ammonia are formed

2. 12.0 moles of hydrogen are required

3. 2.0 moles of nitrogen are required

Explanation:

The balanced chemical equation is:

$N_2+3H_2\rightarrow 2NH_3$

According to stoichiometry:

3 moles of hydrogen form = 2 moles of ammonia

Thus 4.0 moles of hydrogen form = $\frac{2}{3}\times 4.0=2.7moles$ of ammonia

According to stoichiometry:

2 moles of ammonia are formed by = 3 moles of hydrogen

Thus 8.0 moles of ammonia are formed by = $\frac{3}{2}\times 8.0=12.0moles$ of hydrogen

According to stoichiometry:

3 moles of hydrogen react with = 1 mole of nitrogen

Thus 6.0 moles of hydrogen react with = $\frac{1}{3}\times 6.0=2.0moles$ of nitrogen

3 0

2 years ago

1) How much water must be added to 500.mL of 0.200 M HCI to produce a 0.150M solution?

zloy xaker [14]

Answer:

(500.mL)(.200M)/.150M = 667 mL

667mL - 500mL = 167mL of water is needed

1.0 L = 1000mL

M1 V1 = M2 V2

(1.6 mol/L) (175 mL) = (x)(1000mL)

x = .28M

6 0

2 years ago

Explain the connection between reactants, products and limiting reactant.

balu736 [363]

In easy words the connection between Reactants, Products and Limiting reactants is as follow,

Reactants and Products:
Reactants are the starting materials for the synthesis of final synthesized materials called as products.

Example:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

In above reaction Methane (CH₄) and Oxygen (O₂) are the reactants while, CO₂ and H₂O are the products.

Reactants, Products and Limiting Reactants:
Considering the same example it is seen that for one mole of CO₂ two moles of O₂ are required to completely convert into CO₂ and H₂O. If either of the reactant is taken less than the required amount then it will act as a limiting reactant because it will consume first leaving the second reactant present in excess as compare to it. Hence, we can say that the limiting reactant is the starting material which controls the amount of product being formed.

4 0

2 years ago