All categories

3 years ago

How many moles of water molecules are in 57.43 g of iron (iii) chloride dihydrate

Chemistry

1 answer:

Kobotan [32]3 years ago

6 0

Explanation:

im not sure all I know is 1.81 x 1024

You might be interested in

Element X had a mass of 300g and a 1/2 life of 10 years. how many grams will remain after 40 years?

kiruha [24]

<h3>Answer:</h3>

18.75 grams

<h3>Explanation:</h3>

Half-life refers to the time taken by a radioactive material to decay by half of the original mass.
In this case, the half-life of element X is 10 years, which means it takes 10 years for a given mass of the element to decay by half of its original mass.
To calculate the amount that remained after decay we use;

Remaining mass = Original mass × (1/2)^n, where n is the number of half-lives

Number of half-lives = Time for the decay ÷ Half-life

= 40 years ÷ 10 years

= 4

Therefore;

Remaining mass = 300 g × (1/2)⁴

= 300 g × 1/16

= 18.75 g

Hence, a mass of 300 g of an element X decays to 18.75 g after 40 years.

3 0

3 years ago

The elements least likely to form bonds are found in which group? 15, 16, 17, 18

Ipatiy [6.2K]

Answer:

D.18

Explanation:

Edgenuity

3 0

3 years ago

Read 2 more answers

Hello<br>Thank you for not answering my question

julsineya [31]

**Visible confusion** no problem

4 0

3 years ago

How many moles of N2 are need to fill a 35 L tank at standard temperature and pressure?

Tems11 [23]

1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure. Details about moles can be found below.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated using the following formula:

PV = nRT

Where;

P = pressure
V = volume
n = number of moles
R = gas law constant
T = temperature

At STP;

T = 273K
P = 1 atm
R = 0.0821 Latm/molK

1 × 35 = n × 0.0821 × 273

35 = 22.41n

n = 35/22.41

n = 1.56mol

Therefore, 1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure.

Learn more about number of moles at: brainly.com/question/14919968

#SPJ1

4 0

2 years ago

Calculate the percent composition of the following:<br><br> A. VO3<br><br> B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

4 0

3 years ago