Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
When a substance is changing state, its temperature remains constant. This is because energy is used to increase/decrease kinetic energy of the molecules of the substance, increasing/decreasing the inter-molecular distance and overcoming the energy bonds present between the molecules. Therefore, no energy is used to raise the temperature of the substance and therefore it remains constant
Answer:
53.1 mL
Explanation:
Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.
For the ideal gas law:
P1*V1/T1 = P2*V2/T2
Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.
At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:
P1 = 92.5 - 2.8 = 89.7 kPa
T1 = 23°C + 273 = 296 K
89.7*65/296 = 101.325*V2/273
101.325V2 = 5377.45
V2 = 53.1 mL
Answer:
Two hydrogen atoms and one oxygen atom (water) was removed.
Explanation:
yw:))
Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!
