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mario62 [17]
3 years ago
13

A scientist obtains the number 1250.37986 on a calculator. If this number actually has four (4) significant figures, how should

it be written?
A)1251
B)1250.3799
C)1250.4
D)1.250 x 10^3
E)1.250 x 10^–3
Chemistry
2 answers:
8090 [49]3 years ago
5 0

Answer:

D

Explanation:

the zero is trailing with a decimal so it counts. A is wrong because you can't round up, C is too many SF and E is wrong because a negative exponent would give you .001250

Dennis_Churaev [7]3 years ago
4 0

Answer: D

Explanation:

Not A: You don’t round up to 1251 because 3 is less than 5

Not B: That’s more than 4 sig figs

Not C: More than 4 sig figs

D: 1.250 has 4 sig figs and 1250 is the correct answer

Not E: negative exponent makes it into decimal which is way off

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AnnZ [28]

The given question is incomplete. the complete question is:

The world burns the fossil fuel equivalent of approximately 9.50\times 10^{12} kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)

Answer: 29\times 10^{12}kg

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2_C8H_{18}+17O_2\rightarrow 16CO_2+18H_2O

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of octane}=\frac{9.50\times 10^{12}\times 10^3g}{114g/mol}=0.083\times 10^{15}moles

According to stoichiometry :

As 2 moles of octane give = 16 moles of CO_2

Thus 0.083\times 10^{15}moles of octane give =\frac{16}{2}\times 0.083\times 10^{15}=0.664\times 10^{15}moles  of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.664\times 10^{15}moles\times 44g/mol=29.2\times 10^{15}g=29.2\times 10^{12}kg

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5 0
3 years ago
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Answer: A. True

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Answer:

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Explanation:

Hello,

In this case, as the atomic mass of coppe is 63.546 g/mol, with the given mass with can compute the moles as shown below:

mol_{Cu}=848gCu*\frac{1molCu}{63.546 gCu} \\\\mol_{Cu}=13.34molCu

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