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Natali [406]
2 years ago
14

5 points

Chemistry
1 answer:
podryga [215]2 years ago
7 0

Answer:

the answer would be A have good day

Explanation:

pls mark brainliest and 5 stars

<3

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WhT is the amount of NaCI that can be added to 50.0g of water at 60.0oc?
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Answer:

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3 years ago
Calculate the minimum concentration of Mg2+ that mustbe added to 0.10 M NaF in order to initiate a precipitate ofmagnesium fluor
blondinia [14]

Answer:

E. 6.9 E-7 M

Explanation:

  • NaF  →  Na+   +     F-

      0.10M   0.10M    0.10M

  • MgF2 ↔ Mg2+  +    2F-

eq:     S             S          2S + 0.1

∴ Ksp = 6.9 E-9 = [Mg2+][F-]² = (S)(2S+0.1)²

if we compared the concentration ( 0.1 M ) with the Ksp ( 6.9 E-9 ); then we can neglect the solubility as adding:

⇒ Ksp = 6.9 E-9 = (S)(0.1)² = 0.01S

⇒ S = 6.9 E-9 / 0.01 = 6.9 E-7 M

6 0
3 years ago
What can a nation do to improve the exchange rate of its currency?
maksim [4K]

Answer:

A

Explanation:

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4 0
3 years ago
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What happens when the net forces are the same, do you add or subtract it
Strike441 [17]

Answer:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

Explanation:

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8 0
3 years ago
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In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temp
DanielleElmas [232]

<u>Answer:</u> The amount of heat absorbed by the solution is 2.795 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = [70 + 70] = 140 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{140mL}\\\\\text{Mass of water}=(1g/mL\times 140mL)=140g

To calculate the heat absorbed, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 140 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(28.74-23.97)^oC=4.77^oC

Putting values in above equation, we get:

q=140g\times 4.186J/g^oC\times 4.77^oC=2795.4J=2.795kJ

Hence, the amount of heat absorbed by the solution is 2.795 kJ

3 0
3 years ago
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