Ask question

Ask question

All categories

2 years ago

12

Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equati

on.
[ ] SiO2(s) + [ ] HF(g) → [ ] SiF4(g) + [ ] H2O(l)

If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant?

1 answer:

Scrat [10]2 years ago

6 0

Here's your answer, hope this helps!

You might be interested in

You have a substance with a length of 2 cm, width of 3 cm, and a height of 4 cm. Its density is known to be .5 g/cm3. What is th

otez555 [7]

Explanation:

They are related by the the density triangle.

Explanation:

They are related by the the density triangle.

mcdn1.teacherspayteachers.com

d =

m

V

m = d×V

V =

m

d

DENSITY

Density is defined as mass per unit volume.

d =

m

V

Example:

A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?

Step 1: Calculate the volume

V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³

Step 2: Calculate the density

d =

m

V

=

433

g

200

c

m

³

= 2.16 g/cm³

MASS

d =

m

V

We can rearrange this to get the expression for the mass.

m = d×V

Example:

If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?

m = d×V = 500 mL ×

1.11

g

1

m

L

= 555 g

VOLUME

d =

m

V

We can rearrange this to get the expression for the volume.

V =

m

d

Example:

What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.

Step 1: Convert kilograms to grams.

14.83 kg ×

1000

g

1

k

g

= 14 830 g

Step 2: Calculate the volume.

V =

m

d

= 14 830 g ×

1

c

m

³

19.32

g

= 767.6 cm³

5 0

1 year ago

Which best describes a compound such as sodium

tresset_1 [31]

<span>a pure substance that can be separated into different elements by chemical means</span>

3 0

2 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

$\frac{1}{22.4 L}\times 26 L= 1.1607 mol$

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

$\frac{Experimental}{Theoretical}\times 100$

$\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%$

60.42% is the percent yield of the reaction.

8 0

3 years ago

How many degrees of unsaturation are lost or gained in the base-induced transformation from mito-ph to its photoactive ring-open

Fudgin [204]

Unsaturation (IHD) 2 hydrogen Needed

IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)

Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2

The degrees of unsaturation in a molecule are additive — a molecule with one double bond has one degree of unsaturation, a molecule with two double bonds has two degrees of unsaturation, and so forth.

6 0

3 years ago

PLEASE HELP. Give the orbital configuration of the phosphorus (p) atom.

Kipish [7]

Answer: 1s^22s^22p^63s^23p^3

Explanation:

Assuming that orbital configuration is the same as electron configuration this is the answer.

4 0

2 years ago

Read 2 more answers