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3 years ago

C(s) + S8(s) → CS2(l)

1 answer:

4 0

The balanced chemical reaction is written as:

<span>4C(s) + S8(s) → 4CS2(l)

We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.

</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8

The limiting reactant would be S8. We use this amount to calculate.

0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2

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Complite the following reactions. NaOH(aq)+FeBr3(aq)→

stira [4]

Answer:

3NaOH+FeBr3>3NaBr+

Fe(OH)3

Explanation:

After writing the equation it has to be balanced

8 0

3 years ago

If 12.4 mol of Ne gas occupies 122.8 L, how many mol of Ne would occupy 339.2 L under the same temperature and pressure? Record

8090 [49]

Answer:

3.43×10¹ mol

Explanation:

Given data:

Initial number of moles = 12.4 mol

Initial volume = 122.8 L

Final number of moles = ?

Final volume = 339.2 L

Solution:

The number of moles and volume are directly proportional to each other at same temperature and pressure.

V₁/n₁ = V₂/n₂

122.8 L/ 12.4 mol = 339.2 L / n₂

n₂ = 339.2 L× 12.4 mol / 122.8 L

n₂ = 4206.08 L.mol /122.8 L

n₂ = 34.3mol

In scientific notation:

3.43×10¹ mol

7 0

2 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago

How many moles are in 1.05 g of gold (Au)?

Wittaler [7]

Answer:

0.005 mol

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of Gold = 1.05 g ,

m = molecular mass of Gold = 197 g/mol

<u>Hence , moles can be calculated as -</u>

n = w / m = 1.05 g / 197 g/mol = 0.005 mol

7 0

2 years ago

an atom of chlorine has an amount number of 17 and a mass number of 37. how many protons neutrons, and electrons are in the atom

Soloha48 [4]

Answer: There are 20 protons neutrons in the atom

6 0

2 years ago