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lorasvet [3.4K]
3 years ago
8

PLZ HELP I WILL GIVE BRAINLIST ITS DUE IN AN HOUR

Chemistry
1 answer:
FrozenT [24]3 years ago
4 0

oi lmmao where though?

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What is the pressure in atmospheres exerted by a 0.500 mole sample of nitrogen gas in a 10.0 L
Nonamiya [84]

Answer:

The pressure is 1, 22 atm.

Explanation:

We use deal gas formula. First, we convert the unit of temperature in Celsius into Kelvin. We use the constant R= 0,082 l atm /K mol.Then, we solve P (pressure).

0°C=273 K   25°C= 273 + 25= 298 K

PV=nRT   -----> P= (nRT)/V

P= (0,5 mol x 0,082 l atm /K mol x 298 K)/ 10 L

<em>P= 1, 2218 atm</em>

3 0
3 years ago
Can anyone here help me with chemistry
Kay [80]
The answer is 20g N2
5 0
3 years ago
Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th
Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M

The given equilibrium reaction is,

                           N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially conc.      0        0           0.1576

At eqm.               (x)       (x)        (0.1576-2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}

By solving the term, we get:

x=0.0742,0.0839

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

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3 years ago
Your body is much larger now than it was when you were a baby. What is the cause of this growth? *
masya89 [10]

Answer:c

Explanation:

4 0
3 years ago
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Using wolfram alpha or some other reference, determine which of these elements would be liquid at 525 k (assume samples are prot
romanna [79]
The choices for this problem are bismuth, Bi; platinum, Pt; selenium, Se; calcium, Ca and copper, Cu. I think the correct answer would be selenium. The melting point of bismuth is at a temperature of 544.4 Kelvin. At a temperature of 525 K, it would exist as solid. Platinum melts at 2041.1 K. At 525 K, platinum would be in solid form. Selenium has a melting point at 494 K so that at a temperature of 525 K, it would exist in its liquid state. Calcium has a melting point of 1112 K so it would exist as solid at 525 K. Copper has a melting point at 1358 K, so it would still exist as solid at a temperature of 525 K. Therefore, the answer would only be selenium.
3 0
3 years ago
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