All categories

3 years ago

What is the mass of CO2 lost at 20 min from the limestone sample?

Chemistry

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO 3

Hence in sample present CaCO 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO3

You might be interested in

How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?

timama [110]

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

3 0

2 years ago

How many total electrons can enter the set of 3p orbitals? a. 3 b. 10 c. 2 d. 6

Sliva [168]

Answer:

For n=3 and l=1=p

It is 3p-orbital.

Magnetic quantum number m

have values from -l to +l and total of 2l+1 values.

Forl=1, m

values are:

=−1,0,1 for l=1; total m

values =3= Number of orbitals

Each orbital can occupy maximum of two electron

Number of electrons =2×3=6

Thus 6 electrons will show same quantum number values of n=3 and l=1.

Number of elements with last electron in 3p orbitals = 6

5 0

3 years ago

Alkanes are hydrocarbons containing only single bonds. Acyclic alkanes have carbon atoms arranged in chains, whereas cycloalkane

sashaice [31]

Answer:

156 Hydrogen atoms

Explanation:

Any acyclic alkane has a molecular formula that can be expressed as:

CₙH₂ₙ₊₂

Where n is any integer and the number of carbon atoms. For example, Propane has 3 carbon atoms, this means it would have [2*3+2] 8 hydrogen atoms, resulting with a formula of C₃H₈.

An acyclic alkane with 77 carbon atoms would thus have:

2*77 + 2 = 156 hydrogen atoms

4 0

3 years ago

Convert 175 g of H,O into moles.

devlian [24]

Answer:

9.71397613581

that'll help ya

3 0

3 years ago

A corpse discovered in the desert at 5:00 PM was found to have larvae from the blow fly species Phormia regina that had just dev

horrorfan [7]

Answer:

12.19 hours prior to discovery was the corpse placed in the desert.

Explanation:

Heat required for fly eggs to develop into first instar larvae = $Exposure\,time\times temperature$

From the given,

Exposure time = 16 hrs

Temperature = 27 C

$=16\times 27 = 427.2^{o}C$

The day time exposed hours is 10 hrs.-(7.00am-5.00pm)

$=10\times 37.8 = 378^{o}C$

So, the extra heat = 427.2-378= 49.2

The addition heat divided by average temperature.

Average temperature = 22.4 C

$=\frac{49.2}{22.4}= 2.19 hrs$

So, the total time exposed is night time hours to day time hours.

= 10+2.19 = 12.19 hrs.

Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.

3 0

2 years ago