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3 years ago

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Which of the following is not capable of reacting with molecular oxygen?

1 answer:

fredd [130]3 years ago

6 0

I think letter b were you put the letters so

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3 years ago

Describe the formation of ions by electron loss or gain

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Explanation:

Ions form when atoms gain or lose electrons. This is so that they form a full outer shell of electrons. When an atom gains electrons it becomes a negative ion, because electrons are negatively charged. For example, all halogens (group 7 or 17) form negative ions as they gain an electron forming a 1- charge. When an atom loses electrons it becomes a positive ion, as it is losing some negative charge from the electrons. This would be for example, alkali metals (group 1) which lose an electron to form a positive ion with a 1+ charge, (ALL metals form positive ions).

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2 years ago

Which organisms are consumers?

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2 years ago

Read 2 more answers

When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.

EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

<u>Answer:</u> The % yield of the product is 87.05 %

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

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

We are given:

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol$

For the given chemical reaction:

$CaCO_3\rightarrow CaO+CO_2$

By stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of CaO

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of CaO

We know, molar mass of $CaO$ = 56 g/mol

Putting values in above equation, we get:

$\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

$\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%$

Hence, the % yield of the product is 87.05 %

6 0

3 years ago

Choose the atom with:

nata0808 [166]

Answer:

a )Li

b)O

c)F

Explanation:

a) Li-1s^2 2s^1

F-1s^2 2s^2 2p^5

it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy

b)oxygen ion is larger than Na because o have fewer proton

c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.

6 0

3 years ago