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Elenna [48]
3 years ago
9

How many moles of iron (Fe) can be obtained from 3.80 g iron (III) oxide (Fe2O3) reacting with excess carbon monoxide

Chemistry
1 answer:
topjm [15]3 years ago
3 0

0.02375 moles

Explanation:

Fe2O3+CO :. Fe+CO2

We know that

56*2+16*3=160 gm of Fe2O3 can give 56 gm of Fe

So

3.8 gm can give 1.33 gm of Fe

It is equal to 1.33/56 moles of Fe which is equal to 0.02375 moles

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A small windmill is to be built at a science competition. A large floor fan is used to power each windmill The task
fredd [130]

Answer:

D

Explanation:

I'm doing USA test prep and I got it right

7 0
3 years ago
How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
SpyIntel [72]

Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as 5.11 \times 10^{-5} \, s^{-1}

So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

7 0
2 years ago
Read 2 more answers
Iodine monochloride (ICl) has a higher boiling point than bromine (Br2) partly because iodine monochloride is a(n)
tekilochka [14]

Answer: polar molecule.

Explanation:

The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. The boiling point is dependent on the type of forces present.

Iodine monochloride (ICl) is a polar molecule due to the difference in electronegativities of iodine and chlorine. Thus the molecules are bonded by strong dipole dipole forces. Thus a higher temperature is needed to generate enough vapor pressure.

Bromine (Br_2) is a non polar molecule as there is no electronegativity difference between two bromine atoms. The molecules are bonded by weak vanderwaal forces and thus has low boiling point.

7 0
2 years ago
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as
nikitadnepr [17]

The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

Moles of copper(II) nitrate = \frac{469 g}{187.5 g/mol}=2.5013 mol

Moles of sodium sulfide = \frac{156 g}{78 g/mol}=2 mol

According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

Then 2 moles of sodium sulfide will react with:

\frac{1}{1}\times 2mol= 2 mol of copper (II) nitrate

As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.

According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

\frac{2}{1}\times 2mol=4 mol sodium nitrate

Mass of 4 moles of sodium nitrate :

85 g/mol × 4 mol = 340 g

Theoretical yield of sodium nitrate = 340 g

The theoretical yield of sodium nitrate is 340 grams.

7 0
3 years ago
Read 2 more answers
Determine the percent composition for each of the elements in magnesium nitrate, Mg(NO3)2. Please round to the nearest whole num
Verizon [17]

Answer:

% Mg = 16.2 % ≈ 16 %

%N = 18.9 % ≈ 19 %

% O = 64.9 % ≈ 65 %

Explanation:

Step 1: Data given

Molar mass Mg(NO3)2 = 148 g/mol

Molar mass Mg = 24 g/mol

Molar mass N = 14 g/mol

Molar mass O = 16 g/mol

Step 2: Calculate the percent composition

% Mg = (24/148) *100%

% Mg = 16.2 % ≈ 16 %

% N = (2*14 / 148) * 100%

%N = 18.9 % ≈ 19 %

% O = (6*16/148) * 100%

% O = 64.9 % ≈ 65 %

8 0
3 years ago
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