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wel
3 years ago
5

What do all iron atoms and ions have in common ?

Chemistry
2 answers:
aleksley [76]3 years ago
7 0

Answer:

All atoms and ions of iron have 26 protons. Iron's atomic number is 26. Atoms don't gain or lose protons. The atomic mass of this ion is 56, which is the sum of its protons and neutrons.

Phantasy [73]3 years ago
3 0

Answer:

They both have 26 protons.

The one thing ions and iron atoms have in common is that they both have 26 protons. Hope it helps!

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What is the oxidation state of each element in the compound CaS04? Include +or - in your answer as appropriate
borishaifa [10]

Answer:therefore the oxidation state of sulphur is +6 in the compound CaSO4.

Explanation:

5 0
3 years ago
Pleas help with 2 and 4 for brainliest
Snezhnost [94]

mass of pentane : = 30.303 g

moles of Al₂(CO₃)₃ : = 0.147

<h3>Further explanation</h3>

Given

1. Reaction

C₅H₁₂+8O₂→6H₂O+5CO₂.

45.3 g water

2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂

37.2 MgCO₃

Required

mass of pentane

moles of Al₂(CO₃)₃

Solution

1. mol water = 45.3 : 18 g/mol = 2.52

From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :

= 1/6 x mol H₂O

= 1/6 x 2.52

= 0.42

Mass pentane :

= mol x MW

= 0.42 x 72.15 g/mol

= 30.303 g

2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44

mol Al₂(CO₃)₃ :

= 1/3 x mol MgCO₃

= 1/3 x 0.44

= 0.147

4 0
2 years ago
How many molecules are there in 79g of Fe2O3? how many atoms is this?
Leya [2.2K]
There are approximately 160 grams in 1 mol of Fe2O3 molecules. Therefore, there would be 79/160= 0.49375 mols of Fe2O3 molecules in 79 grams. There are 5 atoms in total for each molecule of Fe2O3, therefore 79/160 * 5 = 79/32 = 2.46875 mols of atoms.
4 0
3 years ago
Nitrogen gas consists of _____.
Anni [7]
Nitrogen gas consists of a diatomic nitrogen atom. So, the correct answer is B.
8 0
3 years ago
Read 2 more answers
If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?
myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30.17}\ year^{-1}

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Initial concentration [A_0] = ?

Final concentration [A_t] = 12.5 grams

Applying in the above equation, we get that:-

12.5\ g=[A_0]e^{-0.02297\times 90.6}

[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g

<u>Mass of original sample = 100 g</u>

8 0
3 years ago
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