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2 years ago

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The number 0.0020 expressed in scientific notation is?

2 answers:

Feliz [49]2 years ago

8 0

Answer: Here's your answer: Therefore, the decimal number 0.0020 written in scientific notation is 2 × 10-3 and it has 2 significant figures. Here are some more examples of decimal to scientific notation 0.00200 in scientific notation

Explanation: Pls mark me brainiest pls

Ahat [919]2 years ago

4 0

Answer:

2x $10^{-3}$ (The "x" is the multiplication sign)

Explanation:

A scientific notation is expressed as

$10^{n}$

When you have decimals, the exponent becomes negative:

$10^{-2}=\frac{1}{100} =0.01$

So this is:

2x $10^{-3}$

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Help I need to show work!

lubasha [3.4K]

Answer:

a. Combustion.

b. $m_C=3.057gC$

c. $m_H=3.67gH$

Explanation:

Hello.

a. In this case, given the reaction by which propane is converted into carbon dioxide and water:

$C_3H_8+5O_2 \rightarrow 3CO_2+4H_2O$

It is known as a combustion reaction since it is about a fuel (here propane) which is burnt by the oxygen contained in the air (21%).

b. Since the carbon dioxide contains all the carbon in the products based on the law of conservation of mass, the yielded grams are computed via a mole mass relationship:

$m_C=11.21gCO_2*\frac{12gC}{44gCO_2}=3.057gC$

Since 44 grams of carbon dioxide contain 12 grams of carbon.

c. As well as b., all the hydrogen is given off in form of water, thus, the required mass turns out:

$m_H=33.0gH_2O*\frac{2gH}{18gH_2O}=3.67gH$

Since 18 grams of water contain 2 grams of hydrogen.

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the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

<u>Answer:</u> The entropy change of the process is $2.0\times 10^2J/K$

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

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3 years ago