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Troyanec [42]
3 years ago
15

If you diluted 120 mL of 1,049 M NaCl to 1,920 mL, what would

Chemistry
1 answer:
Alborosie3 years ago
7 0

Answer:

65.56 M

Explanation:

Using the formula as follows:

C1V1 = C2V2

Where;

C1 = initial concentration (M)

C2 = final concentration (M)

V1 = initial volume (L)

V2 = final volume (L)

According to the information in this question,

C1 = 1,049 M

C2 = ?

V1 = 120 mL

V2 = 1,920 mL

Using C1V1 = C2V2

1049 × 120 = C2 × 1920

125880 = 1920C2

C2 = 125880 ÷ 1920

C2 = 65.56 M

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liraira [26]

Answer:

• One mole of oxygen is equivalent to 16 grams.

→ But at STP, 22.4 dm³ are occupied by 1 mole.

{ \tt{22.4 \:  {dm}^{3}   \: \dashrightarrow \: 16 \: grams}} \\  { \tt{33.5 \:  {dm}^{3}  \:  \dashrightarrow \: ( \frac{33.5 \times 16}{22.4} ) \: grams}} \\  \\  \dashrightarrow \: { \boxed{ \tt{23.94 \: g \approx \: 24 \: grams}}}

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2 years ago
Convert a pressure of 0.0248 mm Hg to the equivalent pressure in pascals (Pa). How did you get your answer ?
Rudik [331]
Hope this helps :) remember your conversions and just practice it's fairly easy:)

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2 years ago
What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl
fenix001 [56]
M(P)=3.72 g
M(P)=31 g/mol

m(Cl)=21.28 g
M(Cl)=35.5 g/mol

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n(P)=3.72/31=0.12 mol

n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol

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6 0
3 years ago
If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?
Alex73 [517]

Answer:

0.189 g.

Explanation:

  • This problem is an application on <em>Henry's law.</em>
  • Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
  • Solubility of the gas ∝ partial pressure
  • If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

  • So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0
3 years ago
Describe how temperature changes the average Molecular kinetic energy?
Gwar [14]
An increase in temperature will increase the average kinetic energy of the molecules. As the particles move faster, they will likely hit the edge of the container more often.
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2 years ago
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