Can I please know What are the answers for #s 1,2,3,4 ?

Svetach [21]

Answer:

Francium (Fr)

Explanation:

From the given choices, francium will have the lowest ionization energy.

Ionization energy is the energy required to remove the most loosely held electron within an atom.

The magnitude of the ionization energy depends on the characteristics of the atom in relation to its nuclear charge, atomic radius, stability etc.

Generally on the periodic table, ionization energy increases from left to right on the table
As you go from metals to non-metals and to gases, the value of the ionization energy increases steadily.
Down the group, the value reduces.
Since Francium is the most metallic of all the given choices, it has the highest ionization energy.

3 0

3 years ago

Help needed ASAP, I will mark your answer as brainliest.

beks73 [17]

Answer:

D

Explanation:

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6 0

3 years ago

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8. Bring the balloon in contact with the wall. What happens to the charges in the wall?

Ne4ueva [31]

When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.

Explanation:

Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.

The balloon now as more electrons than normal and therefore has an overall negative charge. Two balloons like this will repel each other.

The other material now has an overall positive charge. Because opposite charges attract, the balloon will now appear to stick to the other material. If you didn't rub the balloon first, it's charge would be neutral and it wouldn't stick to the wall.

7 0

4 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago

True of False: Only conclusions that support the hypothesis should be reported in scientific research.

RSB [31]

False, If something doesn't support your hypothesis maybe it was wrong in the first place and you should make it known in your research if you come up with a different conclusion.

5 0

3 years ago

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