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LuckyWell [14K]
2 years ago
5

Which statement is true about trends in metallic character?a) Metallic character increases as you go to the right across a row i

n the periodic table and as you go down a column.b) Metallic character decreases as you go to the right across a row in the periodic table and increases as you go down a column.c) Metallic character decreases as you go to the right across right across a row in the periodic table and decreases as you go down a columnd) Metallic characterdecreases as you go to the right across a row in the periodic table and increases as you go down a column
Chemistry
1 answer:
klasskru [66]2 years ago
6 0

Answer:

d) Metallic character decreases as you go to the right across a row in the periodic table and increases as you go down a column.

Explanation:

  • <em><u>Metallic character decreases as you move across a period in the periodic table from left to right.</u></em> This is because the attraction between valence electron and the nucleus increases, making it difficult for loss of electrons. It occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell.
  • On the other hand, <em><u>Metallic character increases as you move down an element group in the periodic table.</u></em> This is due to the increase in atomic radius as the number of energy levels increases.
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A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t
Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 2 atm
  • T1= 50 C= 323 K (being 0 C= 273 K)
  • P2= 3.2 atm
  • T2= ?

Replacing:

\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}

Solving:

T2*\frac{2 atm}{323 K} =3.2 atm

T2=3.2 atm*\frac{323 K}{2 atm}

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

5 0
3 years ago
What is the % composition of Carbon in Chromium (iii) Carbonate
photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

\begin{gathered} C\rightarrow3\times12=36 \\  \\ O\rightarrow9\times16=144 \\  \\ Cr\rightarrow2\times52=104 \end{gathered}

The molar mass will be thus:

M=36+104+144=284\text{ g/mol}

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

The percent composition of Carbon is thus 12.7 %.

8 0
1 year ago
TRIPLE POINTS!!! I NEED HELP!!! PLEASE EXPLAIN TOO!!! What is the energy of a quantum of light with a frequency of 4.31 x 1014 1
Lelechka [254]

Answer:

The answer is

2.86 \times  {10}^{ - 19}  \: J

Explanation:

The energy of a quantum of light can be found by using the formula

<h3>E = hf</h3>

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question

f = 4.31 × 10¹⁴ Hz

We have

E = 4.31 × 10¹⁴ × 6.626 × 10-³⁴

We have the final answer as

2.86 \times  {10}^{ - 19}  \: J

Hope this helps you

3 0
3 years ago
A certain electrolyte solution contains 1 gram of salt for every 8 grams of sugar and every 200 grams of water. If the sugar to
12345 [234]

Answer:

The resulting solution contains approximately 666 g of water.

Explanation:

In the initial solution we have:

1g salt : 8g sugar : 200g water

This means that the ratios are:

\frac{salt}{sugar}  = \frac{1}{8} \\\\\frac{sugar}{water} = \frac{8}{200} =\frac{1}{25}

In the final solution we have:

5g salt: xg sugar: yg water

The new ratios are:

\frac{salt}{sugar} = \frac{3}{8} \\\\\frac{sugar}{water} = \frac{1}{50}

Now we can calculate the amount of sugar in the final solution:

\frac{salt}{sugar}  = \frac{5}{x} =\frac{3}{8} \\\\X = 13.333 g

Finally, we calculate the amount of water:

\frac{sugar}{water} = \frac{13.333}{y} = \frac{1}{50} \\y = 666.667 g

4 0
3 years ago
A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba2+. When the concentration of F- exceeds ________ M, BaF
BARSIC [14]

Answer:

When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.

Explanation:

Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)

When BaF₂ precipitates, the Ksp relation is given by

Ksp = [Ba²⁺] [F⁻]²

[Ba²⁺] = 0.0144 M

[F⁻] = ?

Ksp = (1.7 × 10⁻⁶)

1.7 × 10⁻⁶ = (0.0144) [F⁻]²

[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555

[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M

Hope this Helps!!!

7 0
3 years ago
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