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3 years ago

The goal of this lesson is to answer the lesson

Chemistry

2 answers:

OverLord2011 [107]3 years ago

5 0

Answer:

The half life for radioactive can be calculated as:

N /N0 = (1 /2) ^ n

n = T /T half

According to question there are n number of half life are present which would result in remaining amount of element as n.

Explanation:

Sorry if wrong

Scilla [17]3 years ago

4 0

Answer:

1. atom

2. 0.5n

Explanation:

Because half of the atom's nuclei will decay during each cycle.

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1. Name three obstacles that Mendeleev faced in his life?

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Answer:

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2 years ago

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

harkovskaia [24]

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

8 0

3 years ago

An important reaction that takes place in a blast furnace during the production of iron is the formation of iron metal and CO2 f

ladessa [460]

Answer: Mass of $Fe_2O_3$ required to form 930 kg of iron is 1328 kg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For iron:

Given mass of iron = 930 kg = 930000 g (1kg=1000g)

Molar mass of iron = 56 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron}=\frac{930000g}{56g/mol}=16607mol$

The chemical equation for the production of iron follows:

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

By Stoichiometry of the reaction:

2 moles of iron are produced by = 1 mole of $Fe_2O_3$

So, 16607 moles of iron will be produced by = $\frac{1}{2}\times 16607=8303moles$ of $Fe_2O_3$

Now, calculating the mass of $Fe_2O_3$ from equation 1, we get:

Mass of $Fe_2O_3$ = $moles\times {\text {molar mass}}=8303\times 160=1328480g=1328kg$

Thus mass of $Fe_2O_3$ required to form 930 kg of iron is 1328 kg

6 0

3 years ago