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exis [7]
2 years ago
11

Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl2) in 0.30 L of solution. SHOW YOUR

WORK for credit. Partial credit is awarded! Correct answers with no work will not receive credit. Reminder of the the steps: 1. Identify what you know and what you want, and the units. Known: 0.30 L, 44.0 g CaCl2 Want: molarity (mol /L ) 2. Find the amount of CaCl2 in moles. molar Mass of CaCl2 = 110.98 g/mol. Formula: Moles = grams / molar mass 3. Use dimensional analysis to set up and solve equation. Formula: M = mol solute / L solution
Chemistry
1 answer:
ycow [4]2 years ago
6 0

Answer:

[CaCl₂] = 1.32 M

Explanation:

We know the volume of solution → 0.30 L

We know the mass of solute → 44 g of CaCl₂

Let's convert the mass of solute to moles.

44 g . 1 mol / 110.98 g = 0.396 moles

Molarity (mol/L) → 0.396 mol / 0.3 L  = 1.32 M

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How many molecules are there in a 1.43 mole sample of H2O2?
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How many 250 mg tablets of metronidazole are needed to make 150 mL of suspension containing
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How many 250 mg tablets of metronidazole are needed to make 150 mL of suspension containing

100 mg/mL?

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b. 30

c.50

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7 0
2 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
2 years ago
What do all atoms of the same element have
nordsb [41]

Answer:

All atoms of the same element have always have the same amount of protons.

Explanation:

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7 0
3 years ago
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