Since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.
Now we are asked to find the amount concentration of calcium ions and acetate ions in a 0.80 mol/L solution of calcium acetate. The formula of calcium acetate is Ca(CH3COO)2.
Thus;
Ca(CH3COO)2(aq) ----> Ca^2+(aq) + 2CH3COO^-(aq)
It then follows that since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
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Answer:
1.645 moles of excess reactant that is of magnesium metal are left over.
Explanation:
Moles of magnesium metal = 3.29 mol
Moles of HCl = 3.29 mol
According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :
of magnesium metal
Moles of HCl left = 3.29mol - 3.29 mol = 0
Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol
1.645 moles of excess reactant that is of magnesium metal are left over.
Density =
volume = (4.5 cm) * (5.2cm) * (6.0cm)
= 140.40 cm³
since Mass = 1591 g
then Density =
= 11.33 g/cm³
Answer:
Waters' boiling point decreases with increases in elevation because of the atmospheric pressure.
For example, the higher in elevation you are. The lower the atmospheric pressure is. In other words heated water reached boiling point quicker.
Answer:
12.33 cal/sec
Explanation:
As we know,
1 Kcal = 1000 cal
So,
0.74 Kcal = X cal
Solving for X,
X = (0.74 Kcal × 1000 cal) ÷ 1 Kcal
X = 740 cal
Also we know that,
1 Minute = 60 Seconds
Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,
= 740 cal / 60 sec
= 12.33 cal/sec