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3 years ago

Please someone help me out thank you I’ll give brainlist :))

Chemistry

1 answer:

Tom [10]3 years ago

4 0

Answer:

Ionization energy increases going left to right across a period and increases from bottom to top in a group

Electron affinity increases when going up a group

If we are excluding noble gases (aka group 8/18), Chlorine is the element that has the greatest electron affinity. This is because Fluorine's 2p orbital is limited and packed which doesn't quite allow sharing of the orbital with extra electrons easily, while Chlorine has a 3p orbital allowing more space for electrons, where the orbital electrons would be inclined to do so.

Helium is the element with the greatest ionization energy since it's at the top and energy (from Oganesson to Helium) increases when going across a period (from Hydrogen to Helium).

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How many kJ of heat are released by the reaction of 25.0 g of Na2O2(s) in the following reaction? (M = 78.0 g/mol for Na2O2)

solong [7]

-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.

Explanation:

Given:

mass of Na2O2 = 25 grams

atomic mass of Na2O2 = 78 gram/mole

number of mole = $\frac{mass}{atomic mass of 1 mole}$

= $\frac{25}{78}$

=0. 32 moles

The balanced equation for the reaction:

2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ

It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.

similarly 0.3 moles of Na2O2 on reaction would give:

$\frac{126}{2}$ = $\frac{x}{0.32}$

x = $\frac{126 x 0.32}{2}$

= -20.16 KJ

Thus, - 20.16 KJ of energy will be released.

6 0

3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

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

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

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

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

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 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

3 years ago

In the laboratory you dilute 5.32 mL of a concentrated 6.00 M perchloric acid solution to a total volume of 100 mL. What is the

8_murik_8 [283]

Answer:

0.3192 M

Explanation:

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

Volume of stock solution (V1) = 5.32 mL Molarity of stock solution (M1) = 6 M

Volume of diluted solution (V2) = 100 mL

Molarity of diluted solution (M2) =?

We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:

M1V1 = M2V2

6 × 5.32 = M2 ×100

31.92 = M2 × 100

Divide both side by 100

M2 = 31.92 / 100

M2 = 0.3192 M

Therefore, the molarity of the diluted solution is 0.3192 M.

7 0

3 years ago

Which is a set of valid quantum numbers?

Mariana [72]

The last set which is n=4 l=3 m=3 is a valid set

7 0

3 years ago

Read 2 more answers

Calculate the density of a piece of wood, in the units of g/mL, with a volume of 2.5L and a mass of 12.5 kg

Ganezh [65]

Answer:

5 g / ml

Explanation:

Convert the values given to g and ml

12.5 kg = 12500 g

2.5 L = 2500 ml

12500 g / 2500 ml = 5 g/ml

5 0

1 year ago