Answer:
Na2SO4 means: two moles sodium (45.98 g), one mole sulfur (32.06 g), and four moles oxygen (64.00 g) combine to form one mole of sodium sulfate (142.04 g).
Explanation:
The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
Answer:
0.0432M
Explanation:
We begin by writing a balanced equation for the reaction. This is illustrated below:
NaOH + HCl —> NaCl + H2O
From the equation above,
The number of mole of the acid (nA) = 1
The number of mole of the base (nB) = 1
Data obtained from the question include:
Vb (volume of the base) = 54mL
Cb (concentration of the base) = 0.1M
Va (volume of the acid) = 125mL
Ca ( concentration of the acid) =?
Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:
CaVa/CbVb = nA/nB
Ca x 125 / 0.1 x 54 = 1
Cross multiply to express in linear form:
Ca x 125 = 0.1 x 54
Divide both side by 125
Ca = (0.1 x 54) / 125
Ca = 0.0432M
Therefore, the concentration of the acid is 0.0432M
