PLEASEE HELPP ME WITH THIS QUESTIONN!!!!!1

How many moles are there in 3.612 x 1024 formula units of Na2SO4?

vitfil [10]

Answer:

<h3>Theanswer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

Hope this helps you

5 0

3 years ago

What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

kari74 [83]

The heat of reaction (i.e. combustion) of butane ( $C_{4} H_{10}$ ) when reacted with oxygen ( $O_{2}$ )  is -2658 kJ/mol butane, and the chemical reaction is given by:

$C_{4} H_{10}$ + $\frac{13}{2} O_{2}$ ---> $4 CO_{2}$ + $5 H_{2}O$

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane

The mass of carbon dioxide ( $CO_{2}$ ) can be determined by multiplying the moles of butane ( $C_{4} H_{10}$ ) with the mole ratio of ( $CO_{2}$ ) produced to the ( $C_{4} H_{10}$ ) reacted, and then with the molar mass of ( $CO_{2}$ ), which is 44 grams/mole.

Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles $CO_{2}$ / 1 mole $C_{4} H_{10}$ ] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced
= 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.

4 0

3 years ago

Read 2 more answers

The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

adell [148]

Answer:

$1.21x10^{-3}$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^{-3}$

H = $\frac{0.21}{1.38x10^{-3} }$

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = $1.21x10^{-3}$ M

7 0

3 years ago

What is the property of a neutron?

UNO [17]

Answer:

A) 0 charge

Explanation:

neutron=neutral

7 0

3 years ago

Aspirin that has been stored for a long time may give a vinegar like odour and give a purple colour with FeCl3. What reaction wo

iVinArrow [24]

Answer:

See explanation

Explanation:

The IUPAC name of aspirin is 2-Acetoxybenzoic acid. It is composed of an acetoxy moiety and a benzoic acid moiety.

The compound can be hydrolysed under prolonged storage conditions to yield acetic acid which causes the vinegar like odour.

Also, one of the products of this hydrolysis bears a phenol group which reacts with FeCl3 to give a purple color.

3 0

3 years ago