Answer:
31395 J
Explanation:
Given data:
mass of water = 150 g
Initial temperature = 25 °C
Final temperature = 75 °C
Energy absorbed = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 75 °C - 25 °C
ΔT = 50 °C
now we will put the values in formula
q = m . c . ΔT
q = 150 g × 4.186 J/g.°C × 50 °C
q = 31395 J
so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .
Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M
The answer is, 159.69 g/mol
Answer:
see explanation below
Explanation:
In this case, we have the following reaction:
Mg + 1/2O₂ -------> MgO
Now, according to this reaction we want to know the percent composition of MgO. The problem is not providing the mass of the initial reactants and the product, so we can use the atomic weights of the components, to do this
The molecular weight of Mg is 24.305 g/mol, and O is 15.999 g/mol, so, let's calculate the molar mass of MgO:
MM MgO = 24.305 + 15.999 = 40.304 g/mol
Now with this weight, let's see the percent composition of this compound:
%Mg = 24.305 / 40.304 * 100 = 60.304 %
%O = 15.999 / 40.304 * 100 = 39.696 %
And this would be the percent composition of MgO
