There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
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I think The answer to two is it would sink
Answer:
120 g of NaCl in 300 g H20 at 90 C
Explanation:
At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40
we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g
Answer:
1.20 M
Explanation:
Convert grams of Na₂CO₃ to moles. (50.84 g)/(105.99 g/mol) = 0.4797 mol
Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M
The volume of HCl required is 23.89 mL
Calculation of volume:
The reaction:

As HCl and NaOH react in 1 : 1 ratio.
Volume of NaOH= 718 mL
Concentration= 0.183M
Volume of HCl= ?
Concentration= 5.50M
Using the dilution formula:


Therefore,
Volume of HCl required will be 23.89 mL.
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