Which of the following is a step in the conversion of 2.5 hm3 to L?

Write a balanced equation for the reaction of Al(H2O)63+ in aqueous KF. Include the physical states of each reactant and product

olga2289 [7]

Explanation:

The potassium fluoride will dissociate into potassium ions and fluoride ions in their aqueous solution.

$KF(aq)\rightarrow K^+(aq)+F^-(aq)$

So, when 1 mol of hexaaqua aluminium (III) reacts with 6 moles of fluoride ion it gives 1 mole of hexafluoroaluminate(III).

The reaction is given as:

$Al(H_2O)_6^{3+}(aq)+6F^-(aq)\rightleftharpoons AlF_6^{3-}(aq)+6H_2O(l)$

3 0

3 years ago

Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0.250 kilograms, what should be t

maks197457 [2]

The tension should be 0.25

3 0

3 years ago

Read 2 more answers

What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention

Elena L [17]

Answer: Lead(II) nitrate but idk the rest

Explanation:

5 0

2 years ago

In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters

IrinaK [193]

Answer: The volume of hydrogen gas collected over water is 2.13 L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

$\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol$

For the given chemical reaction:

$Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)$

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = $\frac{1}[1}\times 0.0851=0.0851mol$ of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Putting values in above equation, we get:

$733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L$

Hence, the volume of hydrogen gas collected over water is 2.13 L

5 0

3 years ago

A sample of oxygen occupies 20.1 liters under a pressure of 1520 torr at 25.0o What volume would it occupy at 25.0oC if the pres

Zolol [24]

Answer:

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

Explanation:

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Considering an initial state 1 and a final state 2, it is true:

P1* V1= P2*V2

In this case:

P1= 20.1 L
V1= 1520 torr
P2= 760 torr
V2= ?

Replacing:

20.1 L* 1520 torr= 760 torr* V2

Solving:

$V2=\frac{20.1 L* 1520 torr}{760 torr}$

V2= 40.2 L

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

4 0

2 years ago