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3 years ago

The water table is the...

Chemistry

1 answer:

vovangra [49]3 years ago

3 0

B upper boundary of the zone of saturation, land beneath this line is saturated with water.

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Where does the energy that causes evaporation come from?

MissTica

Im pretty sure its the sun.

Hope you have a grrreat day!

4 0

3 years ago

Read 2 more answers

What is the mass of 3.95 l of methane gas at 0.870 atm and 35.0 degrees celsius?

dalvyx [7]

Answer:

2.174 gm

Explanation:

PV = nRT n = number of moles

R = gas constant = .082057 L-atm/(K-mol)

T must be in units of K

.870 (3.95) = n (.082057)(35+273.15)

solve for n = .1359 moles

Methane mole weight (CH4) = 16 gm / mole

.1359 moles * 16 gm/mole = 2.174 gm

5 0

2 years ago

3. A 650 mg sample is analyzed and found to contain

JulsSmile [24]

Answer:

7.85% of Hydrogen in the sample.

Explanation:

51/650 = .07846 x 100= 7.846% (round up)

6 0

4 years ago

Sodium chlorate, NaClO3, decomposes when heated to yield sodium chloride and oxygen, a reaction used to provide oxygen for the e

netineya [11]

Sodium chlorate decomposes into sodium chloride and oxygen.
The reaction is
NaClO3 --Δ--> NaCl + O2
The Δ means that heat is used
This equation is still not balanced.
Balancing the equation
2NaClO3 --Δ--> 2NaCl + 3O2
The equation is now balanced.

3 0

3 years ago

Suppose a 250 ml flask is filled with 1.3mol of O2 and 1.5 mol of NO. The following reaction becomes possible: The equilibrium c

bonufazy [111]

<u>Answer:</u> The equilibrium molarity of oxygen gas is 7.1 M

<u>Explanation:</u>

We are given:

Moles of oxygen gas = 1.3 mole

Moles of NO = 1.5 moles

Volume of the flask = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL)

To calculate the molarity, we use the equation:

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

Molarity of oxygen gas = $\frac{1.3}{0.25}=5.2M$

Molarity of NO = $\frac{1.5}{0.25}=6.0M$

The chemical equation follows:

$O_2+N_2\rightleftharpoons 2NO$

Initial: 5.2 - 6.0

At eqllm: 5.2+x +x 6.0-2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO]}{[O_2][N_2]}$

$K_c=0.394$ (Assuming)

Putting values in above equation, we get:

$0.394=\frac{(6.0-2x)^2}{(5.2+x)\times x}\\\\-3.606x^2+26.0488x-36=0\\\\x=1.9,5.4$

Neglecting the value of x = 5.4 because this cannot be greater than the initial value.

Concentration of oxygen gas at equilibrium = (5.2 + x) = 5.2 + 1.9 = 7.1 M

Hence, the equilibrium molarity of oxygen gas is 7.1 M

8 0

3 years ago