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oksian1 [2.3K]
3 years ago
11

Find the mole ratio of H2SO4 and H20 in the equation Fe2O3 + H2SO4 → Fe2(SO4)3 + H20.

Chemistry
1 answer:
eduard3 years ago
7 0

Explanation:

Fe2O3 +3 H2SO4 → Fe2(SO4)3 + 3H20.

Therefore the ratio is 3:3

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How many grams of Carbon are in sample of 400 grams of Co2?​
RoseWind [281]

Answer:

Mass of carbon = 109.1 g

Explanation:

Given data:

Mass of carbondioxide = 400 g

Mass of carbon = ?

Solution:

Molar mass of carbon = 12 g/mol

Molar mass of CO₂ = 44 g/mol

Mass of carbon in 400g of  CO₂:

Mass of carbon = 12 g/mol/44 g/mol × 400 g

Mass of carbon = 109.1 g

3 0
4 years ago
I need help plss asap
Basile [38]

Answer:

Volume = mass/density

Rearrange the equation for Mass:

Mass = Volume x Density

That is one way you can do it

Conceptually, just look at the units, the wood block's density is 0.6<u>g/cm^3</u> while the volume is 2.2 <u>cm^3</u>

So if density is every gram per centimeter cubed, and the volume is at centimeter cubed, the logical thing to do would be to multiply the density by the volume to get the total mass.

0.6g/cm^3   x   2.2cm^3 = 1.32g

<u>Therefore the mass of the block of wood is 1.32g </u>

8 0
2 years ago
A chemist is given a sample of the CuSO4 hydrate and asked to determine its the empirical formula. The original sample weighed 4
Firlakuza [10]

Answer: The formula for this hydrate is CuSO_4.5H_2O

Explanation:

Decomposition of hydrated copper sulphate is given by:

CuSO_4.xH_2O\rightarrow CuSO_4+xH_2O

Molar mass of CuSO_4 = 160 g/mol

According to stoichiometry:

(160+18x) g of CuSO_4.xH_2O decomposes to give 160 g of anhydrous CuSO_4

Thus 42.75  g of CuSO_4.xH_2O decomposes to give=\frac{160}{160+18x}\times 42.75g of anhydrous CuSO_4

\frac{160}{160+18x}\times 42.75=27.38

x=5

Thus the formula for this hydrate is CuSO_4.5H_2O

8 0
3 years ago
Transition elements have variable valency why​
Anika [276]

Answer:

The transition elements have their valence electrons in two different sets of orbitals that are (n-1)d and ns. Thus, transition elements show variable valencies due to the involvement of penultimate d shell electrons.

Explanation:

6 0
4 years ago
Calculate the values of ΔU, ΔH, and ΔS for the following process:
kykrilka [37]

The values of the changes are

ΔH = 46.44kJ

Δu =  43.34Kj

ΔS = 126.6 J/K

<h3>How to find change in H</h3>

= 1 mol + 75.3 (100 + 273) - 25 + 273 + 1 * 40.79

= 5.6475 + 40.79

= 46.44kJ

<h3>How to find change in S</h3>

1 mol + 75.3 x ln 373/298 + 1 mol x 40.79

= 0.1263

ΔS = 126.6 J/K

<h3>How to find the change in U</h3>

46.44 - 0.00831 * 373

= 43.34Kj

Read more on molar heat here:

brainly.com/question/13439286

#SPJ1

5 0
2 years ago
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