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pashok25 [27]
3 years ago
12

The less mass in a given volume of air the

Chemistry
2 answers:
DiKsa [7]3 years ago
7 0

Answer:

The less mass in a given volume of air the  less dense the air is going to be.

julia-pushkina [17]3 years ago
7 0

Answer:

less dense the air

Explanation:

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What is the bond energy required to break one mole of carbon-carbon bonds​
ArbitrLikvidat [17]

Answer:

<h2><u><em>100 kcal of bond energy</em></u></h2>

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3 years ago
Carbon has an atomic number of 6. This means it has 6 protons and 6 electrons. How many valence electrons does carbon have?
Varvara68 [4.7K]

Answer:

4

Explanation:

Carbon configuration- 2,4

Valence electrons means the outershell electrons

That means valence electrons=4

8 0
3 years ago
Freshwater is distributed in both time and space
Neko [114]
Answer:

Unevenly

Explanation:

Fresh water is distributed unevenly in both time and space.
7 0
2 years ago
Balance the equations by inserting coefficients as needed.
ASHA 777 [7]

Answer:

1.     CaCO3 + 2HCl → CaCl2 + H2O + CO2

2.    C6H12O2 + 8O2 → 6CO2 + 6H2O

Explanation:

7 0
3 years ago
Read 2 more answers
Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

 T_c=1181.55k

Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

3 0
2 years ago
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