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3 years ago

Which of these statements regarding ELECTROLYSIS is FALSE? (see attached photograph)

Chemistry

1 answer:

beks73 [17]3 years ago

4 0

Answer:

Explanation:

Oxidation only occur at Anode and Reduction at Cathode

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The carbon atom has a total of _________ electrons. A.6 B.4 C.2 D.8

denis23 [38]

Carbon has A. 6 atoms

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3 years ago

Read 2 more answers

Balance this chemical equation.

Natasha_Volkova [10]

Answer:

2 C2H6 + 7 O2 = 4 CO2 + 6 H2O

Explanation:

The Bold Numbers are what you should put. This is balanced

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3 years ago

A volume of 25cm3 of a carbonate solution of concentration 0.2mol dm-3 was neutralized by 20 cm3 of acid of concentration 0.5 mo

Galina-37 [17]

Answer: 1 mol of carbonate to 2 mol of acid

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$

a) $\text{Moles of carbonate}=\frac{0.2moldm^{-3}\times 25cm^3}{1000}=0.005mol$

b) $\text{Moles of acid}=\frac{0.5moldm^{-3}\times 20cm^3}{1000}=0.01mol$

Thus the mole ratio of carbonate to acid is = $\frac{0.005}{0.01}=\frac{1}{2}$

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3 years ago

What hold's the nucleus together in an atom?

Simora [160]

The nuclear force holds the nucleus together in an atom

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4 years ago

10) How many grams are there in 1.00 x 10^24 molecules of BC13?

lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0

4 years ago