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3 years ago

How many grams of sulfur must be burned to give 100.0 g of So2

Chemistry

1 answer:

andriy [413]3 years ago

4 0

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) → SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

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Gravity on Earth is 9.8 m/s2, and gravity on the Moon is 1.6 m/s2.

Ivanshal [37]

Answer:

Mass is the same but it weights 64 Newtons

Explanation:

First of Mass is the same in any sort of gravity. Now let's calculate weight

W = MG

where W = Weight

M = Mass

G = Gravity

W = (40kg)(1.6)

W = 64

Sorry for the spelling mistakes, hope this helps

7 0

3 years ago

Read 2 more answers

7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp:

marta [7]

Answer:

4.03dm³

Explanation:

The reaction expression is given as:

3H₂ + N₂ → 2NH₃

Volume of hydrogen = 12dm³

AT rtp:

1 mole of gas occupies volume of 22.4dm³

x mole of hydrogen will occupy a volume of 12dm³

Number of moles of hydrogen = $\frac{12}{22.4}$ = 0.54mole

From the balanced reaction equation:

3 mole of hydrogen gas combines with 1 mole of Nitrogen gas

0.54 mole of hydrogen as will therefore combine with $\frac{0.54}{3}$ = 0.18moles of nitrogen gas

Since ;

1 mole of gas occupies a volume of 22.4dm³

0.18moles of Nitrogen gas will occupy 0.18 x 22.4 = 4.03dm³

6 0

3 years ago

How many seconds are there in 2500 years? (Assume 1 year = 365 days.)

Assoli18 [71]

7.884×10^10 seconds.

4 0

3 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

PLeeease help im so tired and i need to SLEEP. Answer the 1s you know

Bas_tet [7]

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3 0

3 years ago