Answer:
The heat required to change 25.0 g of water from solid ice to liquid water at 0°C is 8350 J
Explanation:
The parameters given are
The temperature of the solid water = 0°C
The heat of fusion, = 334 J/g
The heat of vaporization, = 2260 J/g
Mass of the solid water = 25.0 g
We note that the heat required to change a solid to a liquid is the heat of fusion, from which we have the formula for heat fusion is given as follows;
ΔH = m ×
Therefore, we have;
ΔH = 25 g × 334 J/g = 8350 J
Which gives the heat required to change 25.0 g of water from solid ice to liquid water at 0°C as 8350 J.
Answer:
B
Explanation
Salt water is a solution where the salt completely dissolves therefore is is homogeneous. Solvent is the liquid (water) that dissolves the solute (salt).
The answer is C. Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)
The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
The thermal energy (Q) needed : = 1882.8 J
<h3>Further explanation</h3>
Given
15g water from 25 °C to 50 °C
Required
The thermal energy (Q)
Solution
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Input the value :
Q = 15 g x 4.184 J/g°C x (50 °C- 20 °C)
Q = 1882.8 J
