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1 year ago

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A 42.0 mL sample of an NaOH solution is placed in a flask with a few drops of phenolphthalein (indicator). If 11.7 mL of a 0.325

M HCl solution is needed to reach the endpoint, what is the molarity of the NaOH solution?
NaOH(aq) + HCl(aq) H2O(l) + NaCl(aq)

1 answer:

olganol [36]1 year ago

8 0

Please like if you get the answer :)

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How many grams of CaF2 would be needed to produce 1.12 moles of F2?

zlopas [31]

Step 1 : Write balanced chemical equation.

CaF₂ can be converted to F₂ in 2 steps. The reactions are mentioned below.

I] $CaF_{2} + H_{2} SO_{4} -------> 2HF + CaSO_{4}$

II] $2HF -------> H_{2} + F_{2}$

The final balanced equation for this reaction can be written as

$CaF_{2} + H_{2} SO_{4} ---------> CaSO_{4} + H_{2} + F_{2}$

Step 2: Find moles of CaF₂ Using balanced equation

We have 1.12 mol F₂

The mole ratio of CaF₂ and F₂ is 1:1

$1.12mol F_{2} * \frac{1molCaF_{2}}{1molF_{2}} = 1.12molCaF_{2}$

Step 3 : Calculate molar mass of CaF2.

Molar mass of CaF₂ can be calculated by adding atomic masses of Ca and F

Molar mass of CaF₂ = Ca + 2 (F)

Molar mass of CaF₂ = 40.08 + 18.998 = 78.08 g

Step 4 : Find grams of CaF₂

Grams of CaF₂ = $1.12molCaF_{2} *\frac{78.08gCaF_{2}}{1mol CaF_{2}}$

Grams of CaF₂ = 87.45 g

87.45 grams of CaF2 would be needed to produce 1.12 moles of F2.

4 0

2 years ago

Read 2 more answers

Which of the following substances best conducts electricity in an aqueous solution? CH3OH

lapo4ka [179]

Answer:

KCI

Explanation:

I took the test

5 0

2 years ago

Unless otherwise instructed, you may refer to the periodic table in the Chemistry: Problems and Solutions book for this question

FromTheMoon [43]

Answer:dlobay

Explanation:dncidw

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2 years ago

AleksandrR [38]

Answer:

The mass of the object would be 491.55g/mL. Hope this helped! :)

Explanation:

D= 7.25g/mL M=D•V

V= 67.8mL

M=7.25g/mL•67.8mL

M= 491.55g

8 0

2 years ago

Alex

Answer: 0.714 moles of $PbI_2$ will be produced from 237.1 g of potassium iodide

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of potassium iodide}=\frac{237.1g}{166g/mol}=1.428moles$

The balanced chemical reaction is:

$Pb(NO_3)_2+2KI\rightarrow PbI_2+2KNO_3$

According to stoichiometry :

2 moles of $KI$ produce = 1 mole of $PbI_2$

Thus moles of $KI$ will require= $\frac{1}{2}\times 1.428=0.714moles$ of $PbI_2$

Thus 0.714 moles of $PbI_2$ will be produced from 237.1 g of potassium iodide

6 0

2 years ago

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