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3 years ago

Covert to scientific notation.

Chemistry

1 answer:

gtnhenbr [62]3 years ago

8 0

Answer:

19) 3.6920 × 10⁻⁵

20) 4.059× 10²

21) 2.46810121416 × 10³

22) 1.0000× 10⁻³

Explanation:

Scientific notation is the way to express the large value in short form.

The number in scientific notation have two parts.

The digits (decimal point will place after first digit)

× 10 ( the power which put the decimal point where it should be)

For example:

0.000036920

In scientific notation = 3.6920 × 10⁻⁵

405.9

In scientific notation = 4.059× 10²

2468.10121416

In scientific notation = 2.46810121416 × 10³

0.0010000

In scientific notation = 1.0000× 10⁻³

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alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

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Answer:

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Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?

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Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

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What mass of iron is formed when 10 grams of carbon react with 80 grams of iron iii oxide?

yKpoI14uk [10]

Answer:

55.85 grams of Fe is formed.

Explanation:

Identify the reaction:

2Fe₂O₃ + 3C → 4Fe + 3CO₂

Identify the limiting reactant, previously determine the mol of each reactant

(mass / molar mass)

10 g / 12 g/m = 0.83 moles C

80 g / 159.7 g /m = 0.500 moles Fe₂O₃

2 moles of oxide need 3 moles of C, to react

0.5 moles of oxide, will need ( 0.5 . 3)/ 2 = 0.751 mol

I have 0.83 moles of C, so C is the excess.

The limiting is the oxide.

3 mol of C need 2 mol of oxide to react

0.83 mol of C, will need (0.83 . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)

Ratio is 2:4 (double)

If I have 0.5 moles of oxide, I will produce the double, in the reaction.

1 mol of Fe, will be produce so its mass is 55.85 g

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