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4 years ago

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

2 answers:

7 0

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!

Effectus [21]4 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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3 years ago

17) How many grams are in 0.02 moles of beryllium iodide, Bel2?

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Explanation:

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3 years ago

Read 2 more answers

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

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Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

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Rate in terms of disappearance of $O_2$ = $-\frac{1d[O_2]}{dt}$

Rate in terms of appearance of $NO_2$ = $\frac{1d[NO_2]}{2dt}$

1. The rate of formation of $NO_2$

$-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}$

$\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s$

2. The rate of disappearance of $O_2$

$-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}$

$-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

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3 years ago

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