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3 years ago

Western regions Facts

1 answer:

7 0

This region has both the state with the smallest population (Wyoming) and the largest population (California).
Alaska's land area is much bigger than any other state, but it has one of the lowest populations.
The western US takes up about half of the land of the United States.

The western United States is the region comprising the westernmost states of the United States. As American settlement in the U.S. expanded westward, the meaning of the term the West changed. Before about 1800, the crest of the Appalachian Mountains was seen as the western frontier.

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Hot water is poured into four cups made of different materials (foam, glass, metal, plastic).After one minute, which will feel t

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The answer to this is metal

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If 1.9 kJ of heat is transferred to 96 g aluminum at 113°C, what would the

erastova [34]

Answer:

T2 = 135.1°C

Explanation:

Given data:

Mass of water = 96 g

Initial temperature = 113°C

Final temperature = ?

Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)

Specific heat capacity of aluminium = 0.897 j/g.°C

Solution:

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

ΔT = T2 - T1

Now we will put the values in formula.

Q = m.c. ΔT

1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C

1900 j = 86.112 j/°C × T2 - 113°C

1900 j / 86.112 j/°C = T2 - 113°C

22.1°C + 113°C = T2

T2 = 135.1°C

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3 years ago

0.055 grams of PbSO4 is dissolved in 200.00 grams of H2O.

Greeley [361]

Answer:

Solution's mass = 200.055 g

[PbSO₄] = 275 ppm

Explanation:

Solute mass = 0.055 g of lead(II) sulfate

Solvent mass = 200 g of water

Solution mass = Solvent mass + Solution mass

0.055 g + 200 g = 200.055 g

ppm = μg of solute / g of solution

We convert the mass of solute from g to μg

0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg

5.5×10⁴μg / 200.055 g = 275 ppm

ppm can also be determined as mg of solute / kg of solution

It is important that the relation is 1×10⁻⁶

Let's verify: 0.055 g = 55 mg

200.055 g = 0.200055 kg

55 mg / 0.200055 kg = 275 ppm

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3 years ago

What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?

Andreyy89

Answer: The volume of 0.235 M $H_2SO_4$ needed to titrate 40.0 mL of 0.0500 M $Na_2CO_3$ is 8.51 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Na_2CO_3$

We are given:

$n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL$

Putting values in above equation, we get:

$2\times 0.235\times V_1=2\times 0.0500\times 40.0\\\\V_1=8.51mL$

Thus volume of 0.235 M $H_2SO_4$ needed to titrate 40.0 mL of 0.0500 M $Na_2CO_3$ is 8.51 ml

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Answer:

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Explanation:

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