Answer:
The percent by mass of water in this crystal is:
Explanation:
This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:
- 6,235 g = 100%
- 4.90 g = X
And the value of the variable X is found:
- X = (4.90 g * 100%) / 6,235 g
- X = approximately 78.6%.
The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:
- Water percentage = Total percentage - Percentage after heating.
- <u>Water percentage = 100% - 78.6% = 21.4%</u>
Answer:
The attractive force between them decreases
Explanation:
This is because they become localised.
P = 11.133 atm (purple)
T = -236.733 °C(yellow)
n = 0.174 mol(red)
<h3>Further explanation </h3>
Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:
- Boyle's law at constant T, P = 1 / V
- Charles's law, at constant P, V = T
- Avogadro's law, at constant P and T, V = n
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written

where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
To choose the formula used, we refer to the data provided
Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT
T= 10 +273.15 = 373.15 K
V=5.5 L
n=2 mol

V=8.3 L
P=1.8 atm
n=5 mol

T = 12 + 273.15 = 285.15 K
V=3.4 L
P=1.2 atm

B. 1, 1, 1, 2
Explanation:
You only need to balance the NaNO3 on the right. Since there is 2 NO3 on the left, you need to put a 2 in front of the NaNO3 on the right. Everything else is already balanced so the only coefficient needed is 2 in front of the NaNO3.