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Setler [38]
3 years ago
14

Determine the heat energy needed to raise the temperature of 120 grams of ice at -5 to steam at 115°

Chemistry
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

Q = 30355.2 J

Explanation:

Given data:

Mass of ice = 120 g

Initial temperature = -5°C

Final temperature = 115°C

Energy required = ?

Solution:

Specific heat capacity of ice is = 2.108 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT = T2 -T1

ΔT = 115 - (-5°C)

ΔT = 120 °C

Q = 120 g × 2.108 j/g.°C × 120 °C

Q = 30355.2 J

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A 650.0 g hard-water sample contains 101 mg Ca.what is this concentration in parts per million
ki77a [65]

Answer:

= 155 ppm

Explanation:

PPM also refers to parts per million, it represents a low concentration of a solution. It represents 0.001 gram or a milligram in a 1000 mL, equivalent to 1 mg per liter

Given that;

101 mg of Ca in 650.0 g of water

1 ppm = 1 mg/L

650 g = 650 mL = 0.65 L

Therefore;

= 101 mg/ 0.65 L

=  155.38 mg/L

<u>= 155 ppm</u>

8 0
2 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
The final volume of the solution is 284 mL. What is the concentration of CuSO4 in the final solution, in mol/L?
DochEvi [55]

Answer:

0.0252mol/L

Explanation:

The following data were obtained obtained from the question:

Volume of solution = 284mL = 284/1000 = 0.284L

Mole of CuSO4 = 7.157 × 10^-3 mol

Molarity =?

Molarity = mole/Volume

Molarity = 7.157x10^-3 /0.284

Molarity = 0.0252mol/L

The concentration of the solution is 0.0252mol/L

7 0
3 years ago
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IT SHALL BE C B)!!!!
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7. Which two substances do geologists use in radiocarbon dating?
Sav [38]

Carbon -14 and Carbon 12 are the two substances geologists use in radiocarbon dating.

Answer: Option B

<u>Explanation: </u>

Radiocarbon dating denotes the determination process of the age of fossils of plants or animals based on the ratio of carbon atoms 14 to 12. Carbon naturally exists in two non-radioactive isotopes, Carbon-12 and Carbon-13 and one radioactive isotope carbon 14. The carbon 14 gets released on continuous cosmic reaction with atmospheric nitrogen.

These carbon 14 will be absorbed by the living plants and from the plants. Then, it will enter inside the animals which consume the plants. But once the plants and animals died, they ceased to intake carbon-14. In their living state, the ratios of carbon atoms 14 to 12 in them tends to similar to the ratio in atmosphere.

But after they die, the ratio of C-14 to C-12 will be varying from the ratio of C-14 to C-12 in atmosphere as the concentration of C-14 will be decreasing in the dead animals and plants. Thus using this ratio, geologists can find the fossil's age.

7 0
3 years ago
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