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2 years ago

Tips for high school

Chemistry

2 answers:

Ahat [919]2 years ago

7 0

Answer:

study your tail off, chemistry requires it. I failed because I did not study.

Explanation:

musickatia [10]2 years ago

5 0

Answer:

stoichiometry is hard but the rest is kinda easy if u pay attention

Explanation:

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Which is a physical change?

ira [324]

Making toast is a physical change.

*** Please mark this answer as the Brainliest and leave a Thanks if I helped you! :) ***

4 0

3 years ago

Read 2 more answers

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of alumin

Lerok [7]

Answer: The percentage yield of HF is 73.36 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = $\frac{2}{1}\times 80.05=160.1mol$ of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

$160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg$

To calculate the percentage yield of hydrofluoric acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%$

Hence, the percentage yield of HF is 73.36 %

4 0

3 years ago

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

8 0

3 years ago

Present value lets us compare dollar values from different time periods.

kramer

True? Not sure what the question is

5 0

3 years ago

2. what volume of 0.0150-m hcl solution is required to titrate 150. ml of a 0.0100-m ca(oh)2 solution? (10 points)

laila [671]

100ml volume of 0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.

Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,

C1 V1 =C2 V2

where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1) HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1) Caoh2 solution.

While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give in the expression we get the volume of the final concentration.

V1= C2 V2/ C1

= 0.0100m . 150ml /0.0150M

= 100ml

The volume of the hcl solution is 100ml.

To learn more about dilution formula please visit:

brainly.com/question/7208939

#SPJ4

3 0

1 year ago