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Margaret [11]
3 years ago
5

Can someone help me with the question in the image. if correct i will mark as brainliest

Chemistry
1 answer:
shtirl [24]3 years ago
6 0

Answer: <em>tree rings- kind of plans, rain fall- ice, </em>

Explanation: it goes together

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DaniilM [7]

Your answer would be Transpiration

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3 years ago
weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
bearhunter [10]

Answer:

da fu

Explanation:

are you okay

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3 years ago
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If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,
Evgesh-ka [11]

Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

\frac{P1V1}{RT1} = \frac{P2V2}{RT2}

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

\frac{1*20}{R*10} = \frac{10*V2}{R*100}

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

\frac{1*20*100}{10*10} = V2

∴ V2 = \frac{2000}{100}

∴ V2 = 20mL

3 0
2 years ago
Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

Mass of calorimeter = 1.3 kg = 1300 g

Cs = 3.41 J/g°C

t₁= 25.5 °C

Required

The final temperature, t₂

Solution

Q = m.Cs.Δt

Q out (combustion of compound) = Q in (calorimeter)

24,000 = 1300 x 3.41 x (t₂-25.5)

t₂ = 30.9 °C

3 0
3 years ago
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20 POINTS TO WHO ANSWERS THIS
Xelga [282]

answer: D

Here is a list of the most common ways to speed up a chemical reaction

Increase the temperature (reactions that absorb energy)

Decrease the temperature (Reactions that release energy)

3 0
3 years ago
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