<span>0.74 Kcal/min x 1000 cal/60 s
</span>
Answer: 0.225 atm
Explanation:
For this problem, we have to use Boyle's Law.
Boyle's Law: P₁V₁=P₂V₂
Since we are asked to find P₂, let's manipulate the equation.
P₂=(P₁V₁)/V₂
With this equation, the liters cancel out and we will be left with atm.
P₂=0.225 atm
From the calculations, the concentration of the acid is 0.24 M.
<h3>What is neutralization?</h3>
The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.
We have to use the formula;
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O
CA = ?
CB = 1.2 M
VA = 50 mL
VB = 20 mL
NA = 1
NB = 2
CA = CBVBNA/VANB
CA = 1.2 M * 20 mL * 1/ 50 mL * 2
CA = 0.24 M
Learn more about neutralization:brainly.com/question/27891712
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The amount
per 100 g is:
38.7 %
calcium = 38.7g Ca / 100g compound = 38.7g
19.9 %
phosphorus = 19.9g P / 100g compound = 19.9g
41.2 %
oxygen = 41.2g O / 100g compound = 41.2g
The molar amounts of calcium,
phosphorus and oxygen in 100g sample are calculated by dividing each element’s
mass by its molar mass:
Ca = 38.7/40.078
= 0.96
P = 19.9/30.97
= 0.64
O = 41.2/15.99
= 2.57
C0efficients
for the tentative empirical formula are derived by dividing each molar amount
by the lesser value that is 0.64 and in this case, after that multiply wih 2.
Ca = 0.96 /
0.64 = 1.5=1.5 x 2 = 3
P = 0.64 /
0.64 = 1 = 1x2= 2
O = 2.57 /
0.64 = 4= 4x2= 8
Since, the
resulting ratio is calcium 3, phosphorus 2 and oxygen 8
<span>So, the
empirical formula of the compound is Ca</span>₃(PO₄)₂
