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balandron [24]
2 years ago
13

8.310x10^2 – 7.210x10^1[?]x10^[?]​

Chemistry
1 answer:
alisha [4.7K]2 years ago
7 0

Answer:

7.589\times 10^{2}

Explanation:

The expression can be solved mathematically as follows:

1) 8.310\times 10^{2}-7.210\times 10^{1} Given

2) 8.310\times 10^{1+1} - 7.210\times 10^{1} Definition of sum

3) (8.310\times 10^{1})\times 10^{1}-7.210\times 10^{1} a^{m+n} = a^{m}\cdot a^{n}/Associative property

4) (8.310\times 10^{1}-7.210)\times 10^{1} Distributive property

5) (83.100-7.210)\times 10^{1} Multiplication

6) 75.89\times 10^{1} Subtraction.

7) (7.589\times 10^{1})\times 10^{1} Multiplication/Associative property

8) 7.589\times (10^{1}\times 10^{1}) Associative property

9) 7.589\times 10^{2} a^{m+n} = a^{m}\cdot a^{n}/Result

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Answer:

the normality of the given solution is 0.0755 N

Explanation:

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Here we have to realize the two sodiums ions per carbonate ion i.e.

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= 0.1886eq ÷ 0.2500L

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2 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

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Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

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The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

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