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Digiron [165]
3 years ago
8

Which of the following reactions substitutes one element in a reactant with another to form a product?

Chemistry
2 answers:
Kruka [31]3 years ago
4 0
I believe the answer is A. single-displacement 
Ann [662]3 years ago
3 0
Its A. because it substituting one element with another. If it said its substituting 2 then it would be double replacement
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Calcium has an atomic number of 20. How many electrons of a calcium atom are involved in chemical bonding?
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The answer is 2 electrons.


The electron configuration of calcium is 2:8:8:2

Calcium has two electrons in its outermost shell. These are its valence electrons and are the ones used in bonding with other elements.  Valence electrons of an atom are those electrons that are in its outer energy shell or that are available for bonding.

Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known  as an ionic compound.

For example, when calcium metal reacts with chlorine gas,   calcium gives up its two valence electrons and Chlorine accepts them resulting in a new substance called calcium chloride in which the two elements have ended up forming ionic bonds.

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Where are the rare earth metals located on the periodic table
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3 years ago
The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We have given in the question

P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

8 0
3 years ago
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