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3 years ago

What is the empirical formula of sulphur dioxide which contains 50% by mass of sulphur

Chemistry

2 answers:

expeople1 [14]3 years ago

7 0

Answer:

Question what are you doing the same thing you guys have a good time

Explanation:

is this the same question why did they do that and how much they do it

RUDIKE [14]3 years ago

4 0

Answer:

Empirical formula is SO2.

Explanation:

So the fraction of Sulphur is 0.50 and the fraction of oxygen is 0.50.

Dividing by the atomic masses of sulphur and oxygen:

S = .50 / 32 = 0.015625

O = .50 / 16 = 0.3125

Ratio of S: O = 0.015625/ 0.015625 to 0.3125/ 0.015625

= 1:2.

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What is 31 grams converted to milligrams

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31,000 milligrams is equal to 31 grams

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(06.04 HC)

Ksju [112]

The mass of sodium chloride at the two parts are mathematically given as

m=10,688.18g
mass of Nacl(m)=39.15g

<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>

Generally, the equation for ideal gas is mathematically given as

PV=nRT

Where the chemical equation is

F2 + 2NaCl → Cl2 + 2NaF

Therefore

1.50x15=m/M *(1.50*0.0821)

1-50 x 15=m/58.5 *(1.50*0.0821)

m=10,688.18g

Part 2

PV=m'/MRT

1*15=m'/58.5*0.0821*273

m'=39.15g

mass of Nacl(m)=m'=39.15g

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2 years ago

If 2.0×10−4 moles of S2O2−8 in 170 mL of solution is consumed in 170 seconds , what is the rate of consumption of S2O2−8?

Ostrovityanka [42]

Answer : The rate of consumption of $S_2O_2^{-8}$ is, $7.0\times 10^{-6}M/s$

Explanation : Given,

Moles of $S_2O_2^{-8}$ = $2.0\times 10^{-4}mol$

Volume of solution = 170 mL = 0.170 L (1 L = 1000 mL)

Time = 170 s

First we have to calculate the concentration.

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume of solution}}$

$\text{Concentration}=\frac{2.0\times 10^{-4}mol}{0.170L}$

$\text{Concentration}=1.2\times 10^{-3}M$

Now we have to calculate the rate of consumption.

$\text{Rate of consumption}=\frac{\text{Concentration}}{\text{Time}}$

$\text{Rate of consumption}=\frac{1.2\times 10^{-3}M}{170s}$

$\text{Rate of consumption}=7.0\times 10^{-6}M/s$

Thus, the rate of consumption of $S_2O_2^{-8}$ is, $7.0\times 10^{-6}M/s$

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3 years ago

What would the products be when aluminum chloride (which contains Al3+ and Cl- ions) is melted and electrolyzed? Write half equa

Hoochie [10]

The product of electrolysis : Al at cathode and Cl₂(chlorine) at anode

<h3>Further explanation</h3>

Given

Aluminum chloride compound

Required

The product of electrolysis

Solution

The rule :

The reaction at the cathode(the negative pole) :

1. the reduced active metal is water, other than that the metal will be reduced

2. H⁺ of the acid will be reduced

The reaction at the anode((the positive pole) :

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

In the electrolysis of molten AlCl₃ with an inert electrode, the cation will be reduced at the cathode and the anion will be oxidized at the anode

Cathode : Al³⁺ + 3e⁻ ⇒ Al(s)

Anode : 2Cl⁻⇒Cl₂(g) + 2e⁻

6 0

3 years ago

What is the empirical formula of sulphur dioxide which contains 50% by mass of sulphur​

What is the empirical formula of sulphur dioxide which contains 50% by mass of sulphur