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3 years ago

13

La lluvia acida es un fenomeno a nivel local o Regional?

1 answer:

Lesechka [4]3 years ago

4 0

Answer:

Regional

Explanation:

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you rent a car in Germany with a gas mileage rating of 12.8km/L. What is its rating in miles per gallon?

Oksi-84 [34.3K]

so one liter is about 2. something gallons so at 12.8 km/l you would get about 30 mpg which is insane but yeah

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3 years ago

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Describe how the size of the sun compares to the sizes of the other stars.

alina1380 [7]

Answer:

The sun's circumference is about 2,713,406 miles (4,366,813 km). It may be the biggest thing in this neighborhood, but the sun is just average compared to other stars. Betelgeuse, a red giant, is about 700 times bigger than the sun and about 14,000 times brighter.

Explanation:

looked it up

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2 years ago

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If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain

skad [1K]

Answer:

11.31 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>

<em></em>

<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>

6 0

3 years ago

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A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

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3 years ago

HELP ME AGAIN FOR BRAINLIEST

oksano4ka [1.4K]

Answer:

1: C

2: a

3: B

4: D

hope this helps

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3 years ago