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Dmitrij [34]
3 years ago
10

Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough

temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt MnSO 4 ( s ) . A 16.260 g sample of the hydrate was heated to 300 ∘ C . The resulting MnSO 4 ( s ) had a mass of 14.527 g . Calculate the value of n in MnSO 4 ⋅ n H 2 O ( s ) .
Chemistry
1 answer:
Mama L [17]3 years ago
8 0

Answer:

Value of n in MnSO₄.nH₂O is one.

Explanation:

The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,

Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)

Mass of H₂O = 16.260 g - 14.527 g

Mass of H₂O = 1.733 g

moles of H₂O = (1.733 g) ÷ (18.015 g/mole)

moles of H₂O =  0.0962

For the moles of anhydrous salt:

moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄

moles of MnSO₄ = 14.5277 ÷ 151.001

moles of MnSO₄= 0.0962

Now for n:

n = moles of water ÷ moles of MnSO₄

n = 0.0962 ÷ 0.0962

n = 1

The above calculations show that one mole of H₂O is attached to the  one formula unit of MnSO₄

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Answer:

In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.

Explanation:

C1×N1 = C2×N2

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3 years ago
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Answer:

\Delta \rho =2.22 g/mL

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL

Best regards.

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I don't know how to do this can I get the answers plz it's due in 1 hour
katrin [286]
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
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Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
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