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Dafna11 [192]
3 years ago
13

a sample of neon occupied a volume of 15.0L at a temp 30.0(C. if the temp were to increase by 15(C, what would be the volume of

the neon under this new condition ?
Chemistry
2 answers:
jeyben [28]3 years ago
4 0

Answer:

22.5L will be the new volume for the gas

Explanation:

Pressure and number of moles remains constant.

When the temperature for a sample of gas is increased, the volume will also be increased, acording to this relation:

V₁ / T₁ = V₂ / T₂

15L / 30°C  =  V₂ / 45°C

(15L / 30°C ). 45°C = 22.5L

We assume Ar, as an Ideal Gas

OLEGan [10]3 years ago
4 0

Answer:

When the temperature increases with 15 °C the new volume will be 15.7 L

Explanation:

Step 1: Data given

Initial volume of the neon = 15.0 L

Initial temperature of the neon = 30.0 °C = 303 K

The temperature is increased by 15°C = 318 K

Step 2: Calculate the new volume

V1/T1 = V2/T2

⇒with V1 = the initial volume of neon = 15.0 L

⇒with T1 = the initial temperature = 303 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the increased temperature = 318 K

15.0 L / 303 K = V2 / 318 K

V2 = (15.0 L/ 303 K) * 318 K

V2 = 15.7 L

When the temperature increases with 15 °C the new volume will be 15.7 L

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2: <u>FALSE</u> Neutrons are the particles in an atom that have a <u>neutral charge</u>. They aren't positive like protons. They aren't negative like electrons.

3: <u>TRUE</u> Dmitri Ivanovich Mendeleev was a Russian chemist and inventor. He is best remembered for formulating the Periodic Law and creating a farsighted version of the periodic table of elements. He is also known as the <u>"father of the periodic table" </u>

6 0
3 years ago
Read 2 more answers
5. What are the coefficients when the equation below is balanced?
aleksley [76]

Answer:

C.) 2,1,2

Explanation:

You have to balance the equation by getting the number of elements on the left and right side to equal. Before balancing there is one more oxygen on the left side than the right side. You have to change the coefficient in front of H2O to fix this, and that means adding a "2" in front of the H2 to get that to equal as well.

You end up with 2H₂+10₂→ 2H₂O

8 0
2 years ago
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

3 0
4 years ago
Can you help me please ​
myrzilka [38]

Answer:

A. Non-metal

B. Six

C. 2

D.NA2O

E.Oxygen

Explanation:

Oxygen is the element in the second period (second horizontal row) and Group VI A of the periodic table. These properties apply to oxygen

7 0
2 years ago
The formula for zirconium(VI) dichromate
Iteru [2.4K]

Answer:

Zr(Cr2O7)2

Explanation:

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3 years ago
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