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vovangra [49]
2 years ago
14

ANSWER FAST PLEASE

Chemistry
1 answer:
sasho [114]2 years ago
6 0
<h2><em><u>2</u></em><em><u>.</u></em><em><u>8</u></em><em><u> </u></em><em><u>J</u></em><em><u> </u></em></h2>

<h2><em><u>this</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>right answer helpful</u></em></h2>

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How is the periodic law demonstrated in halogens
inn [45]
Halogens is defined as the group of 7 periodic table. As, every periodic table contains 7 valence electrons and they only need 1 more to complete an outer shell, that is why they are extremely reactive. And according to the law that recurring patterns of the properties of elements arise when they are arranged in order of increasing atomic number. As the halogen all act very similarly with each other in chemical reaction, it is true.
3 0
2 years ago
Read 2 more answers
C5H12, pentane, is a liquid at room temperature.
Art [367]

Answer:

C)We cannot be sure unless we find out its boiling point.

Explanation:

It is necessary to clearly explain here that simply observing two compounds of the same homologous series irrespective of how close they may be in the series will not give us the faintest idea regarding which one will be a liquid, solid or gas at room temperature.

However, to determine whether an unknown substance will be a liquid at room temperature, then its important to measure its boiling point. If the boiling point is above room temperature, and the melting point is below room temperature, the compound is a liquid. If the boiling point of the unknown substance is below room temperature, it is a gas.

It is now safe to conclude that cannot decide on the state of matter in which a compound exists unless we know something about its boiling point, not merely looking closely at the properties of its neighbouring compounds in the same homologous series

8 0
2 years ago
Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)
Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0
2 years ago
You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
A copper rod that has a mass of 200.0 g has an initial temperature of 20.0°C and is heated to 40.0°C. If 1,540 J of heat are nee
Romashka [77]

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)</span>

1540 = 200.0 (C)(40 - 20)

<span> <span>C = 0.385 J / g C</span></span>

<span><span>
</span></span>

<span><span>Hope this answers the question. Have a nice day.</span></span>

8 0
2 years ago
Read 2 more answers
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