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Alik [6]
3 years ago
10

A 12.5 g sample of granite initially at 82.0 oc is immersed into 25.0 g of water that is initially at 22.0 oc. what is the final

temperature of both substances when they reach thermal equilibrium? (specific heat capacity of water = 4.18 j/g.oc and specific heat capacity of granite = 0.790 j/g.oc)
Chemistry
1 answer:
Paraphin [41]3 years ago
7 0

The amount of heat lost by granite is equal to the amount of heat gained by water. Therefore their change in enthalpies must be equal. The opposite in sign means that one is gaining while the other is losing

ΔH granite = - ΔH water

ΔH is the change in enthalpy experienced by a closed object as it undergoes change in energy. This is expressed mathematically as,

ΔH = m Cp (T2 – T1)

Given this information, we can say that:

12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) = - 25.0 g * 4.18 J / g ˚C * (T2 – 22 ˚C)

9.875 (T2 – 82) = 104.5 (22 – T2)

9.875 T2 – 809.75 = 2299 – 104.5 T2

114.375 T2 = 3108.75

T2 = 27.18 ˚C

The temperature of 2 objects after reaching thermal equilibrium is 27.18 ˚<span>C.</span>

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Biotic and abiotic factors are the environmental conditions that the organisms have to face to live in a specified environment.

-Abiotic factors-
Abiotic factors are the physical and chemical conditions of an environment. For example : heat, salinity, pressure, light, wind, pH ...

-Biotic factors-
Biotic factors are all the biological conditions of an environment for a specie/taxa. It can include prey and predator abundance, available food amount, available space, intra and interspecific competition...

The development of organims is under the control of abiotic factors. Some are adapted to heat, cold etc ... The abiotic factors will define which organisms are able or not to live in a specified place.

The living organisms will constitute the biotic factors, which define if and how can an organism live in a specified environment.
So, the abiotic factors are controling the biotic factors of an environment.

Hope it helps you !
4 0
3 years ago
Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhou
RUDIKE [14]

Answer:

a) 2.541 mol/MJ;

b) 1.124 mol/MJ;

c) 0.4354 mol/MJ;

d) 0.1835 mol/MJ

Explanation:

The enthalpy of formation (ΔH°f) is the enthalpy of a reaction to form a compound by its constituents. For CO₂, ΔH°f = - 393.5 kJ/mol.

The enthalpy of a reaction is the sum of the enthalpy of the products (each one multiplied by the number of moles) less the sum of the enthalpy of the reactants (each one multiplied by the number of moles). The ΔH°f for simple substances (with one atom) is 0. The combustion is the reaction between the fuel and the oxygen.

a) The combution reaction is:

C(s) + O₂(g) → CO₂(g)

ΔH°rxn = -393.5 kJ/mol = -393.5x10⁻³ MJ/mol

Number of moles per MJ released: 1/|ΔH°rxn|

n = 1/(393.5x10⁻³) = 2.541 mol/MJ

b) The combustion reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

H₂O is in the liquid state because it's at 1 atm and 25ºC.

ΔH°f, H₂O(l) = -285.3 kJ/mol

ΔH°f, O₂(g) = 0

ΔH°f, CH₄(g) = -74.8 kJ/mol

ΔH°rxn = [2*(-285.3 ) + 1*(-393.5)] - [1*(-74.8)]

ΔH°rxn = -889.3 kJ/mol = -889.3x10⁻³ MJ/mol

n = 1/889.3x10⁻³ = 1.124 mol/MJ

c) C₃H₈(g) + 10O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔH°f,C₃H₈(g) = -25.2 kJ/mol

ΔH°rxn = [4*(-285.3) + 3*(-393.5)] - [1*(-25.2)]

ΔH°rxn = -2,296.5 kJ/mol = -2.2965 MJ/mol

n = 1/2.2965 = 0.4354 mol/MJ

d) C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(l)

ΔH°f, C₈H₁₈(l) = -250.1 kJ/mol

ΔH°rxn = [9*(-283.5) + 8*(-393.5)] - [1*(-250.1)]

ΔH°rxn = -5,449.4 kJ/mol = -5.4494 MJ/mol

n = 1/5.4494 = 0.1835 mol/MJ

4 0
3 years ago
A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0
2 years ago
The arrangement of things in the order that they occurred.
RoseWind [281]
The answer is B: chronological order
4 0
3 years ago
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Which is NOT a common property of a metal?
xenn [34]

Answer:

maybe it's a

Explanation:

because metal can not bend

6 0
2 years ago
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