Ask question

Ask question

All categories

3 years ago

6

How many grams of calcium phosphate can be produced when 78.5 grams of calcium hydroxide reacts with excess phosphoric acid? Unb

alanced equation: H3PO4 + Ca(OH)2 → H2O + Ca3(PO4)2 So I balanced the equation as 2H3PO4 + 3Ca(OH)2 = 6H2O + Ca3(PO4)2 but now I am stuck

1 answer:

sasho [114]3 years ago

3 0

Once you balance the enquation you "switch partners" of the element (negative charge to positive charge)

You might be interested in

8). How many moles of oxygen gas must be placed in a 35.0 L container in order

klemol [59]

Answer:

4.1 moles

Explanation:

Applying

PV = nRT................ equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT.............. Equation 2

From the question,

Given: V = 35 L , P = 2.8 atm, T = 15 °C = (15+273) = 288 K, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (35×2.8)/(0.083×288)

n = 4.1 moles

5 0

3 years ago

If I initially have a gas at a pressure of 12 kPa, a volume of 23 liters, and a temperature of 200 K, and then I raise the press

sertanlavr [38]

Answer:

The new volume of gas would be 30 L.

Explanation:

This is an example of a Combined Gas Laws problem.

5 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

<u>Answer:</u>

<u>For A:</u> The $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u> The $K_c$ for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

<u>For A:</u>

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

If a gas has a volume of 3.67 L and a pressure of 790 mm Hg, what will the pressure be if the volume is compressed to 2.12 L? Wh

oksian1 [2.3K]

If each gas sample has the same temperature and pressure, which has the greatest volume? Since hydrogen gas has the lowest molar mass of the set, 1 g will have the greatest number of moles and therefore the greatest volume. What is the Ideal Gas Law?

7 0

2 years ago

Read 2 more answers

A 508-g sample of sodium bicarbonate (NaHCO3) contains how many moles of sodium bicarbonate (NaHCO3)?

Blizzard [7]

molar mass = (22.99) + (1.01) + (12.01) + 3(16.00)

molar mass = 84.01 g/mol

//

(508g)(1 mol/84.01 g) = 6.0

There are 6.0 moles of sodium bicarbonate

3 0

3 years ago