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Rudik [331]
3 years ago
11

Which of these contains generic material but is not classified as living?

Chemistry
1 answer:
denis-greek [22]3 years ago
4 0
Viruses are not classified as being alive in because of the fact that they can't reproduce.
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On a balance, you have beakers of AgNO3AgNO3 solution and NaClNaCl solution. When mixed, they will form AgCl(s)AgCl(s). What wil
statuscvo [17]

Explanation:

According to the law of conservation of mass, mass can neither be created nor it can be destroyed. But it can be simply transformed from one form to another.

Therefore, when AgNO_{3} is added to NaCl then the compound formed will have same mass as that of reactants.

       AgNO_{3} + NaCl \rightarrow AgCl + NaNO_{3}

Total mass of reactants is (169.87 + 58.44) g/mol = 228.31 g/mol

Total mass of products is (143.32 + 84.99) g/mol = 228.31 g/mol

Thus, we can conclude that mass of the new mixture will stay the same.

6 0
3 years ago
Potassium is isotopic and has RAM of 39.5 work out the percentage abundance of each isotope in a given sample of potassium which
Troyanec [42]

Yo sup??

Let the percentage of K-39 be x

then the percentage of K-40 is 100-(x+0.01)

We know that the net weight should be 39.5. Therefore we can say

(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5  

(since we are taking it in percent)

39*x+40*(100-(x+0.01))+38*0.01=3950

39x+4000-40x-0.4+0.38=3950

2x=49.98

x=24.99

=25 (approx)

Therefore K-39 is 25% in nature and K-40 is 75% in nature.

Hope this helps.

3 0
3 years ago
The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

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

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

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

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
Please help me with my science thank you!
Tatiana [17]
I cant see it maybe post it again?
4 0
3 years ago
Which of the following is considered a greenhouse gas?
Sergeeva-Olga [200]
The primary greenhouse gases are water vapor, carbon dioxide, methane, nitrous oxide, and ozone. So, the correct answer among these options is water vapor, that is considered to be a greenhouse gas.
4 0
3 years ago
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