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3 years ago

8

9) Which element in each pair has the larger ionization energy?

1 answer:

erastovalidia [21]3 years ago

6 0

I think the answer is C.

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Propose a 3-step reaction for the conversion of cyclopentene to propanedioic acid

Gnom [1K]

I
don't know because this is the question which I never heard

6 0

3 years ago

What type(s) of atoms make up the element carbon?

ivolga24 [154]

A, as carbon is an atomic substance - element

8 0

3 years ago

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Two elements are studied. One has atomic number X and one has atomic number X+2. It is known that element X is a halogen. To wha

vladimir2022 [97]

Answer:

Group 1 (or IA)

Explanation:

If element X is a halogen, then it belongs to the group 17 (or VIIA, under a different notation).

For each extra unit of atomic number, the group number increases by 1. That means that the X+1 element would belong to the group 18 (or VIIIA). <em>The X+2 element would thus belong in the group 1 </em>(or IA) one period higher (higher as in numeric value, not as in position in the periodic table).

7 0

3 years ago

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

= 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

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7 0

2 years ago

Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

5 0

4 years ago

Read 2 more answers