All categories

3 years ago

What mass of Ba3(PO4)2 is contained in 1575 mL of a 0.35 M solution of Ba3(PO4)2?

Chemistry

1 answer:

irina [24]3 years ago

3 0

Answer:

= 331.81 g

Explanation:

Molarity is calculated by the formula;

Molarity = Moles/volume in liters

Therefore;

Moles = Molarity ×Volume in liters

= 0.35 M × 1.575 L

= 0.55125 Moles

But; Molar mass of Ba3(PO4)2 is 601.93 g/mol

Thus;

Mass = 0.55125 moles × 601.93 g/mol

= 331.81 g

You might be interested in

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 ℃, and the temperature af

gtnhenbr [62]

Answer:fH = - 3,255.7 kJ/mol

Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0

3 years ago

2CO (g) + O2 (g) → CO2 (g) How many grams of O2 are needed to react with 6.64 L of CO2?

allsm [11]

,12.3

Dgrdhc xhyh ghcz dhn tub

8 0

3 years ago

100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

3 years ago

Be sure to answer all parts. Calculate the pH of the following two buffer solutions: (a) 1.4 M CH3COONa/1.6 M CH3COOH. (b) 0.1 M

aalyn [17]

Answer:

a) pH = 4.68 (more effective)

b) pH =4.44.

Explanation:

The pH of buffer solution is obtained by Henderson Hassalbalch's equation.

The equation is:

$pH =pKa +log\frac{[salt]}{[acid]}$

a) pKa of acetic acid = 4.74

[salt] = [CH₃COONa] = 1.4 M

[acid] = [CH₃COOH] = 1.6 M

$pH = 4.74 + log \frac{1.4}{1.6}= 4.68$

This is more effective as there is very less difference in the concentration of salt and acid.

b) pKa of acetic acid = 4.74

[salt] = [CH₃COONa] = 0.1 M

[acid] = [CH₃COOH] = 0.2 M

$pH = 4.74 + log \frac{0.1}{0.2}= 4.44$

3 0

4 years ago

One gram of carbon contains 50,200,000,000,000,000,000,000 atoms. express the number of carbon atoms in scientific notation

german

N(C)=5,02·10²² atoms

calculation check:

N(C)=(1/12)*6.022*10²³=0.5018*10²³≈5.02·10²²

5 0

3 years ago