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3 years ago

How many grams of zinc are needed to produce 25.0 g ZnCl2

1 answer:

5 0

To solve this question, you would first need to write the reaction and then be able to convert one unit to another the critical unit of moles and then be able to convert it back into grams.

Zn + Cl2 -> ZnCl2.

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Hydrogen gas is collected over water at 21 degrees Celsius. At 21 degrees Celsius the vapor pressure of water is 18.7 torr. If t

Alinara [238K]

Answer : The pressure of hydrogen gas is, 739.3 torr

Explanation :

As we are given:

Vapor pressure of water = 18.7 torr

Barometric pressure = 758 torr

Now we have to calculate the pressure of hydrogen gas.

Pressure of hydrogen gas = Barometric pressure - Vapor pressure of water

Pressure of hydrogen gas = 758 torr - 18.7 torr

Pressure of hydrogen gas = 739.3 torr

Therefore, the pressure of hydrogen gas is, 739.3 torr

3 0

3 years ago

Reacting Solutions of Aluminum Chloride with Potassium Hydroxide

gizmo_the_mogwai [7]

1Al2(SO4)3 + 3ZnCl2 → 2AlCl3 + 3ZnSO4

The coefficients represents moles. There is 1 mole of Aluminum Sulfate, 3 moles of Zinc(II) Chloride, 2 moles of Aluminum Chloride, and 3 moles of Zinc(II) Sulfate.

Now add all the coefficients/moles.

9 is the sum of all the coefficients.

7 0

2 years ago

Read 2 more answers

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho

ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= $\frac{59}{760}atm$

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

$n = \frac{PV}{RT}$

$n = \frac{0.078 * 15000}{0.082*290}$

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since numbers of moles = $\frac{mass}{molar mass}$

and mass = density × vollume

Then; we can say ;

number of moles = $\frac{density*volume }{molar mass of ethanol}$

number of moles = $\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}$

number of moles = $\frac{&85}{46.07}$

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0

3 years ago

What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if

Ede4ka [16]

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

Can be formulated:

Reaction: aA ---> bB

$\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}$