Answer:

Explanation:
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In this case, given the chemical reaction, we can compute the grams of oxygen by using the 98.2 g of water via the 2:1 mole ratio between them, the molar mass of water that is 18.02 g/mol, the molar mass of gaseous oxygen that is 32.00 g/mol and the following stoichiometric procedure relating the given information:

In which the result is displayed with three significant figures because the given mass of water 98.2 g, has three significant figures too.
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Answer:
The ΔH of the reaction is + 12.45 KJ/mol
Explanation:
Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)
heat capacity of water = 4.18 Jk-1 Mol-1
Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)
Molar mass of NaHCO3 = 84 g/mol
Mole of NaHCO3 = 14.5 / 84 = 0.173 mol
Step 1 : Calculate the heat energy (Q) lost by the water.
Q = M x C x ΔT
Q = -100 x 4.18 x (-5.14)
Q = 2148.5 joules
Q = 2.1485 K J
Step 2: Calculating the ΔH of the reaction?
ΔH = Q / number of moles of NaHCO3
ΔH = 2.1485 / 0.173
ΔH = 12.42 KJ/mol
Answer:
0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.
Given data:
Molarity = 5.00 M
Formula Mass = 58.5 g/mol
Required volume = 0.800 L
To Find;
Mass in gram = ?
Solution:
Formula for calculating mass in gram is given as,
Mass in gram = Molarity × Formula mass × Volume required / 1000 putting values
Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000
Mass in gram = 0.234 g
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