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marysya [2.9K]
3 years ago
9

determine the specific heat of a mayterial if a 35g sample of the material absorbs 48J as it is heated from 298K to 313K

Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
6 0
Data:

m = 35g
Q = 48 j
T1 = 298K
T2 = 313K
Cs = ?

Fórmula: Q = m*Cs*(T2 - T1) => Cs = Q / [m (T2 - T1)]

Cs = 48 j / [35g (313K - 298K] = 0.0914 j / g*K = 91.4 j / kg*K
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For the reaction: N2(g) + H2(g) → NH3(g)
gayaneshka [121]

The rate of increase of NH3 is 0.22M/s.

<h3>What is the balanced equation of the reaction?</h3>

The balanced equation of the reaction is given below:

  • N2(g) + 3 H2(g) → 2 NH3(g)

The rate of decrease of N2 is half the rate of increase of NH3.

Rate of decrease of N2 = -0.11 M/s

Rate of increase of NH3 = 2 × +0.11 M/s = 0.22M/s.

In conclusion, the rate of formation of products is dependent on the rate of disappearance of reactants.

Learn more about rate of reaction at: brainly.com/question/25724512

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3 0
2 years ago
A 2.5 mol sample of phosphorus pentachloride, PCl5 dissociates at 160C and 1.00atm to give 0.338 mol of phosphorus trichloride a
Readme [11.4K]

Explanation:

Moles of phosphorus pentachloride present initially = 2.5 mol

Moles of phosphorus trichloride at equilibrium = 0.338 mol

PCl_5\rightleftharpoons PCl_3+Cl_2

Initially

2.5 mol      0    0

At equilibrium:

(2.5 - x) mol      x     x

So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol

Mass of 0.338 moles of  phosphorus trichloride at equilibrium:

= 0.338 mol × 137.5 g/mol = 46.475 g

Moles of phosphorus pentachloride present at equilibrium :

= (2.5 - 0.338) mol = 2.162 mol

Mass of 2.162 moles of  phosphorus pentachloride at equilibrium:

= 2.162 mol × 208.5 g/mol = 450.777 g

Moles of chloride gas present at equilibrium : 0.338 mol

Mass of 0.338 moles of chloride gas at equilibrium:

= 0.338 mol × 71 g/mol = 23.998 g

3 0
3 years ago
Which of these materials requires the most heat to raise its temperature by one degree
AlladinOne [14]

Answer:

Water ______________________________________________

3 0
2 years ago
Uses of carbon dioxide.​
andrey2020 [161]

Answer:

hello can you follow me

Explanation:

used in fire extinction, blasting rubber, foaming rubber and plastic

6 0
2 years ago
How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?
vodomira [7]

Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

Water ⇒ C = 1 cal/g°C

m = 65.8 g

Ti = 31.5°C

Tf = 36.9°C

Heat, Q = ?

2) Formula:

Q = mCΔT

3) Calculations:

Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal

4) You can convert from calories to Joules using the conversion factor:

1 cal = 4.18 J

⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J

3 0
3 years ago
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