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andreyandreev [35.5K]
3 years ago
14

A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place

d in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Chemistry
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

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olga2289 [7]

Answer:

chemestry

Explanation:

5 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
What are the products of the reaction below?:<br> calcium chloride+ silver nitrate
WARRIOR [948]

Answer:

Calcium chloride react with silver nitrate to produce calcium nitrate and silver(I) chloride.

6 0
2 years ago
Read 2 more answers
6. 7. A hyperbaric chamber has a volume of 200. L. (a) How many moles of oxygen are needed to fill the chamber at room temperatu
Otrada [13]

Answer:

a) 24.7 mol

b) 790 g

Explanation:

Step 1: Given data

  • Volume of the chamber (V): 200. L
  • Room temperature (T): 23 °C
  • Pressure of the gas (P): 3.00 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 23°C + 273.15 = 296 K

Step 3: Calculate the moles (n) of oxygen

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol

Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen

The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.

m = n × M

m = 24.7 mol × 32.00 g/mol = 790 g

6 0
3 years ago
In the chemical reaction, how many grams of nitrogen are needed to completely react with 24.3 grams of hydrogen
vlabodo [156]

Answer:

112.540 gram of N2

Explanation:

From the balanced chemical equation

N2+3H2----->2NH3

It means that 1 mole of Nitrogen needs 3 moles of hydrogen

In term of mass,we can say that

6.048 gram of H2 needs 28.01 gram of N2

or, 1 gram of H2 needs (28.01/6.048) gram of N2

and 24.3gram of H2 needs (24.3*28.01/6.048) gram of N2

=112.540 gram of N2

5 0
3 years ago
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