which of the following is an example of an electromagnetic wave A. an ocean wave B. A sound wave C. A light wave D. A wave trave
1 answer:
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Answer:
55.0 g/mol
Step-by-step explanation:
<em>Step 1</em>. Partial pressure of hydrogen
You are collecting the gas over water, so
At 25 °C,
===============
<em>Step 2</em>. Moles of H₂
We can use the <em>Ideal Gas Law</em>.
pV = nRT Divide both sides by RT
n = (pV)/(RT)
p = 732.2 Torr Convert to atmospheres
p = 732.2/760
p = 0.9634 atm
V = 291 mL Convert to litres
V = 0.291 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 25 °C Convert to kelvins
T = (25 + 273.15 ) K = 298.15 K
n = (0.9632 × 0.291)/(0.082 06 × 298.15)
n = 0.2804/24.47
n = 0.011 46 mol
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<em>Step 3</em>. Moles of metal
The partial chemical equation is
M + H₂SO₄ ⟶ H₂ + …
The molar ratio of M:H₂ is 1 mol M:1 mol H₂.
Moles of M = 0.011 46× 1/1
Moles of M = 0.011 46 mol M
===============
<em>Step 4</em>. Atomic mass of M
Atomic mass = mass of M/moles of M
Atomic mass = 0.630/0.011 46
Atomic mass = 55.0 g/mol
Solubility of a compound in water can be referred to as the amount of the compound that can be dissolved in 1 L of the solvent (water) at any given temperature. Solubility of a compound can be expressed in the units of g/L or mg/L.
Given that the solubility of calcium carbonate in water = 14 mg/L
We have to calculate the volume of water that can dissolve 11 g of calcium carbonate.
Converting 11 g calcium carbonate to mg:
Volume of water that would dissolve 11000 mg calcium carbonate
=
=785.7 L
Rounding the volume 785.7 L to two significant figures, we get 790 L water.
Therefore, we would need 790 L water to completely dissolve 11 g of calcium carbonate.
Answer:
The relative humidity is 36%.
Explanation:
Relative humidity is the measure of ratio of actual vapor density to saturated vapor density in a given system. So there are different ways to measure the relative humidity in the atmosphere. One of the common ways is to measure using Psychrometer. In this instrument, two different thermometers are used as measuring device. One of the thermometer measures the humidity in dry air and another in wet air. So the difference of the temperature of dry and wet air bulb will be related to determine the relative humidity percentage. The formula to determine the relative humidity using this method is as below.
Also,
And,
Here e =2.718, and
are the temperature at dry bulb and wet bulb temperatures, respectively.
So, the relative humidity after substituting the values of Td = 2 C and Tw = -2 C is 36%.