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Vikentia [17]
3 years ago
5

When the temperature and number of particles of a gas are constant, which of the following is also constant?

Chemistry
2 answers:
wariber [46]3 years ago
7 0

Answer:

According to Gas Law

Volume is inversely proportional to pressure, if the number of particles and the temperature are constant. There are two ways for the pressure to remain the same as the volume increases.

iogann1982 [59]3 years ago
5 0

Answer:

i dont see the pic

Explanation:

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Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com
lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

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

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30  = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0
3 years ago
3. Give three examples of a pure substances
Nadya [2.5K]

Answer:

There will be weight, there will be volume, there will be height

Explanation:

5 0
3 years ago
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Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g) Which half reaction correctly describes the oxidation that is taking place?
abruzzese [7]
Zn (s) -> Zn+2 (aq) + 2e-

Zn (s) with a neutral charge is oxidized and looses two electrons in the process to form ZnCl2 (aq) where Zn has a charge of 2+.
8 0
3 years ago
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An INPUT of photosynthesis is ______________<br> A.Oxygen<br> B.Roots<br> C.Sunlight
AfilCa [17]

The inputs of photosynthesis are light energy, and matter in the form of water absorbed through the roots, and carbon dioxide absorbed through the leaves.

Therefore an input of photosynthesis from the choices given is,

Sunlight

3 0
3 years ago
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At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti
mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

\frac{R}{0.120}=\frac{1}{4}

R=0.03M/s

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0
3 years ago
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