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3 years ago

I’m confused how to do this

Chemistry

1 answer:

Stels [109]3 years ago

7 0

Answer:

B. P (Phosphorus)

Explanation:

The element is be phosphorus.

This is true because phosphorus has a larger atomic radius than carbon. If you move than the group, elements gain additional electron shell. That additional electron shell keeps electrons far from the nucleus of the atom. This actually increases the atomic radius. Also, phosphorus is more electronegative than aluminum because phosphorus itself is a non-metal and non-metals are generally electronegative while aluminum is a metal and electropositive.

Phosphorus has a lower ionization energy than argon because the ionization energy across the period (i.e from left to right) increases and across period 3, you will find phosphorus first before argon in the periodic table.

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Will give brainliest!!

Nikolay [14]

Answer: True.

Explanation: A teakettle does have a greater frequency than a drumbeat. The teakettle has a higher pitched sound and frequency than a low pitched sound like a drumbeat. So a teakettle is high and a drumbeat is low.

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2 years ago

Read 2 more answers

3. What is the greatest amount of H2O that can be made with 3.8 moles of H and 5 moles of

Pachacha [2.7K]

Answer: Hello!

first i believe we need a balanced equation to start...

i got 2H2 + 1O2 = 2H2O

This tells us that we need 2 moles of H2 for every 1 mole of O2 Since we only have 1 mole of H2 compared to the 5 moles of O2 hydrogen is the limiting reagent. For illustration, divide the balanced equation by 2 in order to get 1 mole of H2 If we start with 1.0 moles of H2 we'll produce 1.0 mole of H2O

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2 years ago

What is the difference between being imaginative in doing science and doing psuedoscience?

Dovator [93]

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3 years ago

What does the pH of a solution indicate

Nina [5.8K]

I think it’s B because I just know

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3 years ago

What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300

Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: $T_{1}$ = 237 K, $P_{1}$ = 101.33 kPa, $V_{1}$ = ?

$T_{2}$ = 300 K, $P_{2}$ = 151.99 kPa, $V_{2}$ = 0.50 L

Formula used is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L$

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

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3 years ago