Answer:
The answer is 0.75M HCl
Explanation:
To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:
To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:
C1V1 = C2V2
C1 = concentration of acid
C2 = concentration of base
V1 = volume of acid
V2 = volume of base
From the information supplied in the question:
concentration of acid (HCl) is the unknown
volume of acid (HCl) = 10ml
concentration of base (NaOH) = 0.15M
volume of base (NaOH) = 50ml
C1 x 10ml = 0.15M x 50ml
C1 x 10 = 7.5
divide both side by 10
C1 = 0.75M
concentration of acid (HCl) is 0.75M
Scientists have known the consequences of mixing sodium bicarbonate and vitamin C ever since 1936, when a study on the subject was published in the "Journal of Nutrition." The authors of this study measured the amount of vitamin C recovered from the urine of people who drank a fixed amount of orange juice. The authors determined that the amount of vitamin C excreted was decreased by administration of sodium bicarbonate. Followup studies in the 1940s showed that this effect was due to the neutralization of the vitamin C by the sodium bicarbonate.
Answer: The balanced equation for the given reaction is
.
Explanation:
A chemical equation which contains same number of atoms on both reactant and product side.
For example, 
Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply
by 2 on reactant side and multiply
by 2. Hence, the equation will be re-written as follows.

Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.
Thus, we can conclude that the balanced equation for the given reaction is
.
Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.