All categories

3 years ago

How many moles of sodium phosphate would be required to react with 1.0 mol of silver nitrate?

1 answer:

7 0

The reaction equation:
Na₃PO₄ + 3AgNO₃ → 3NaNO₃ + Ag₃PO₄

From the equation, we can see that:
1 mole of sodium phosphate requires 3 moles of silver nitrate. So one mole of silver nitrate will require:
1/3 moles of sodium phosphate.

You might be interested in

A gas occupies a volume of 202 ml at a pressure of 505 torr. To what pressure must the gas be subject in order to change the vol

UNO [17]

Answer:

1569 torr

Explanation:

Assuming ideal behaviour and constant temperature, we can solve this problem by using Boyle's law, which states:

V₁P₁ = V₂P₂

Where in this case:

V₁ = 202 mL

P₁ = 505 torr

V₂ = 65.0 mL

P₂ = ?

We input the data given by the problem:

202 mL * 505 torr = 65.0 mL * P₂

And solve for P₂:

P₂ = 1569 torr

6 0

3 years ago

6. Which statement is true about all metals?

timofeeve [1]

Answer:

A They are attracted to a magnet

6 0

3 years ago

Read 2 more answers

How interactions cause substance to boil at different tempurtures

Nata [24]

The surface area is a big help

3 0

3 years ago

Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).

dedylja [7]

The balanced chemical equation between HCl and $Ba(OH)_{2}$ is:

$2HCl (aq) + Ba(OH)_{2}(aq) -->BaCl_{2}(aq) + 2 H_{2}O(l)$

Moles of $Ba(OH)_{2}$ = $1.00 g Ba(OH)_{2} * \frac{1 mol Ba(OH)_{2}}{171.3 g Ba(OH)_{2}} = 0.00584 mol Ba(OH)_{2}$

Moles of HCl required to neutralize $Ba(OH)_{2}$ :

$0.00584 mol Ba(OH)_{2} * \frac{ 2 mol HCl}{1 mol Ba(OH)_{2}} = 0.01168 mol HCl$

Calculating the volume of HCl from moles and molarity:

$0.01168 mol HCl * \frac{1 L}{0.100 mol} * \frac{1000 mL}{1 L} = 116.8 mL$

7 0

3 years ago

Read 2 more answers

A 600.0 mL sample of krypton is held at standard conditions. What volume does the gas occupy when the pressure is tripled and th

solong [7]

Answer: The volume when the pressure and temperature has changed is 400 mL

Explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=P_1\\V_1=600.00mL\\T_1=T_1\\P_2=3P_1\\V_2=?\\T_2=2T_1$

Putting values in above equation, we get:

$\frac{P_1\times 600mL}{T_1}=\frac{3P_1\times V_2}{2T_1}\\\\V_2=\frac{600\times P_1\times 2T_1}{T_1\times 3P_1}=400mL$

Hence, the volume when the pressure and temperature has changed is 400 mL

6 0

3 years ago