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3 years ago

Science- I need help with a question really quick I’m ending school soon

Chemistry

2 answers:

nasty-shy [4]3 years ago

8 0

Bar graph, and charge your phone

miv72 [106K]3 years ago

7 0

The answers bar graph

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What causes the jet stream? (4 points)

mamaluj [8]

The answer is A.

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kwrob

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3 years ago

How does a Camel prevent the loss of water from their body?

Schach [20]

Camels lose less water through their urine and feces than many other mammals. Their kidneys concentrate water heavily, leading to salty urine. The intestines also reabsorb water from intestinal material as it is digested, so dry feces are produced.

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3 years ago

Thorium-234 has a half life of about 25 days. What percent of the thorium will remain after 125 days

solong [7]

Answer:

3.125% will remain after 125 days

Explanation:

Given data:

Half life of Th-234 = 25 days

Percent of thorium remain after 125 days = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 125 days / 25 days

Number of half lives = 5

At time zero =100%

At 1st half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

At third half life = 25%/ 2 = 12.5%

At 4th half life = 12.5% /2 = 6.25%

At 5th half life = 6.25% /2 = 3.125%

4 0

4 years ago

The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo

klio [65]

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

$Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ$

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

$Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ$

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

$Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C$

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

Learn more: brainly.com/question/24245395

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3 years ago

1. Which groups of the periodic table constitute p-block?

DENIUS [597]

Groups 13 through 18 are the constitute p-block

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3 years ago