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olasank [31]
3 years ago
6

Muiltplying 2.5 x 10^10 by 3.5 x 10^-7

Chemistry
1 answer:
andrey2020 [161]3 years ago
6 0

<u>Answer: </u>

The value of \left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right) is 8750

<u>Solution: </u>

\left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right)

\Rightarrow\left(2.5 \times 10^{10} \times 3.5 \times 10^{-7}\right)

\Rightarrow\left(2.5 \times 3.5 \times\left(10^{10} \times 10^{-7}\right)\right)

\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)

\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)

\Rightarrow(2.5 \times 3.5) \times 10^{3}

\Rightarrow 8.75 \times 10^{3}

\Rightarrow 8.75 \times 1000=8750

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1 mole of Ca has a mass of 40 g.

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8 0
3 years ago
. How much energy is lost by a 30.0g sample of water that decreases in temperature from 56.7C to 25.0C?
lbvjy [14]

Answer:

Q = -3980.9 j

Explanation:

Given data:

Mass of sample = 30 g

Initial temperature = 56.7 °C

Final temperature = 25 °C

Specific heat of water = 4.186 j/g.°C

Amount of heat released = ?

Formula:

Q = m.c.ΔT

Q = heat released

m = mass of sample

c = specific heat of given sample

ΔT = change in temperature

Solution:

ΔT = T2 -T1

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Q = m.c.ΔT

Q = 30 g × 4.186 j/g.°C ×  - 31.7°C

Q = -3980.9 j

8 0
3 years ago
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