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EastWind [94]
3 years ago
10

What is Ka for H3PO4(aq) ⇌ H+(aq) + H2PO4-(aq) a. Ka = (H3PO4)/(h+)(H2PO4-) B. Ka = (H3PO4)(H+)(H2PO4-) C. Ka = (H3PO4)/(H+)(H2P

O4-) D. Ka = (H+)(H2PO4-)/(H3PO4)
Chemistry
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

The answer is option d.

Ka = (H+)(H2PO4-)/(H3PO4)

Hope this helps.

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Calculate the concentration of a solution that has 6.7 moles in a volume of 0.6 liters.
Yuri [45]

Answer:

11.2 M

Explanation:

Given data:

Number of moles = 6.7 mol

Volume of solution = 0.6 L

Concentration /Molarity = ?

Solution:

Molarity:

It is number of moles of solute in to per kg or litters of solution. It can be calculated by the following formula.

Molarity = number of moles / Volume in L

Now we will put the values in formula.

Molarity = 6.7 mol / 0.6 L

Molarity = 11.2 mol/L

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6 0
4 years ago
One peanut M&M weighs approximately 2.33 g.
Sphinxa [80]

Answer:

There are 23076 peanut M&M's in 53.768 kg of M&M's.

Explanation:

First we <u>convert 53.768 kg into g</u>:

  • 53.768 kg * 1000 = 53768 g

Then we <u>divide the total mass of M&M's by the mass of one peanut M&M,</u> in order to calculate the answer:

  • 53768 g / 2.33 g = 23076

So there are 23076 peanut M&M's in 53.768 kg of M&M's.

3 0
3 years ago
25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

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