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vampirchik [111]
3 years ago
12

Compute 4.659×104−2.14×104. Round the answer appropriately.

Chemistry
2 answers:
IrinaK [193]3 years ago
7 0
<span>To compute 4.659×104−2.14×104, the first step is the factorization. That is as follows:4.659×104−2.14×104= 10^4.(4.659−2.14), the next step is to compute 4.659−2.14=2.51, so 10^4.(4.659−2.14)=2.51x10^4=2.51x 10000 (because10^4=10000), the last calculus is 2.51x 10000=25100, the final answer is 25,000.Hope this helps. Let me know if you need additional help!</span>
Effectus [21]3 years ago
3 0

Answer is: proper number of significant figures is 25190.00.

10⁴ = 10×10×10×10.

10⁴ = 10000.

1) 4.659×10⁴ = 4.659×10000.

4.659×10⁴ = 46590.

2) 2.14×10⁴ = 2.14×10000.

2.14×10⁴ = 21400.

3) 4.659×10⁴ - 2.14×10⁴ = 46590 - 21400.

4.659×10⁴ - 2.14×10⁴ = 25190 (2.519×10⁴).

An integer is a number that can be written without a fractional component.

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Explanation:

As we know,

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Solving for X,

                      X =  (0.74 Kcal × 1000 cal) ÷ 1 Kcal

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Also we know that,

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The molecular formula of compound is Al(C_{2}H_{3}O_{2})_{3}.

Since, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6.023\times 10^{23} atoms of  Al(C_{2}H_{3}O_{2})_{3}.

Thus, according to molecular formula, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6\times 6.023\times 10^{23}=3.61\times 10^{24} atoms of oxygen atoms.

1 atom of oxygen will be present in \frac{1}{3.61\times 10^{24}} moles of Al(C_{2}H_{3}O_{2})_{3} . Thus,

2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}.

Molar mass of Al(C_{2}H_{3}O_{2})_{3} is 236 g/mol, mass can be calculated as follows:

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