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3 years ago

6

a 55g block of metal has a specific heat of 0.45 J/g°C. What will be the temperature change of this metal if 450 J of heat energ

y are added

1 answer:

Dafna11 [192]3 years ago

7 0

Answer:

18.18C

Explanation: Write whats below to show your work :D

m=55

C=0.45

Q=450zj

ch.temp=?

Q=mct

450 = (55)(.45)T

450 = 24.75

/24.75 = /24.75 (cancels out)

18.18 is the answer

Hope this helps! did it with my teacher.

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How many moles of a gas would occupy 11.4 L at 273K and 2.00 atm?

bagirrra123 [75]

Answer:

1.02mol

Explanation:

Using the general gas equation below;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

P = 2.0 atm

V = 11.4L

T = 273K

n = ?

Using PV = nRT

n = PV/RT

n = 2 × 11.4/ 0.0821 × 273

n = 22.8/22.41

n = 1.017

n = 1.02mol

3 0

3 years ago

Which state of matter would be described as having highly energized, charged particles that move extremely fast

uranmaximum [27]

There are four states of matter, solid, liquid, gas and plasma. Their formation is as when solid is heated it converts into liquid, liquid on heating converts into gases and gases on heating converts into plasma.

Plasma:
Plasma is the fourth state of matter. It is the highest energy state of matter.

Composition:
Plasma is made up of negatively charged and positively charged particles.

Result:
The answer to your question is Plasma.

8 0

3 years ago

Express 2/10 in scientific notation

Novay_Z [31]

The answer would be 2.0 x 10^-1

5 0

4 years ago

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

3 years ago

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

<em>V₂ = 12,31L</em>

I hope it helps!

6 0

4 years ago