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3 years ago

13

How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through

a solution containing CrCl3

1 answer:

True [87]3 years ago

5 0

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

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Your dad is working on creating a brick border for the lake in your backyard each brick has a mass of 100 g and a volume of 20 c

wel

Answer:

5000kg/m³

Explanation:

density=mass/volume

d=m/v

d=100/20

=5g/cm³

g/cm³*1000=kg/m³

5g/m³*1000=5000kg/m³

4 0

3 years ago

Read 2 more answers

Will Florine react? will it combine with other elements?

Alexxx [7]

Fluorine is the most active chemical element, reacting with virtually every element. It even reacts with the noble gases at high temperatures and pressures. The noble gases, or Group 18 (VIIIA), also known as the inert gases, generally do not react with other elements.

Explanation:

5 0

4 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago

What are atoms?

user100 [1]

Answer:

The smallest particle of a chemical element can be defined as an atom.

Explanation:

The number of protons in one atom of an element determines the atom's identity, and the number of electrons determines its electrical charge.

a single electron or one of two or more electrons in the outer shell of an atom that is responsible for the chemical properties of the atom is known as valence electrons.

An atom's reactivity is its tendency to lose or gain electrons. ... This is because they have one outer electron and losing it gives them the stability of a outer electron shell as the next level... The reactivities of elements can be predicted by periodic trends.

7 0

3 years ago

Read 2 more answers

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours

Vanyuwa [196]

(a) Power of bulb is 100 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{100}{1000}kW=0.1 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.1 kW\times 2190 h=219 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

$Cost=\$ (219\times 0.15)=\$ 32.85$

Therefore, annual cost of incandescent light bulb is $\$ 32.85$

(b) Power of bulb is 25 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{25}{1000}kW=0.025 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.025 kW\times 2190 h=54.75 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

$Cost=\$ (54.75\times 0.15)=\$ 8.21$

Therefore, annual cost of fluorescent bulb is $\$ 8.21$ .

7 0

3 years ago