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4 years ago

What do elements in the same group on the periodic table have in common?

Chemistry

1 answer:

Andreyy894 years ago

5 0

Answer:

They have same ionic charge and valence electrons.

Explanation:

This is because, these elements have the increase in nuclrar charge and screening effect.

Therefore, since they have same number of valence electrons, they form ions with the same charge.

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What are you and what do you want with my attendance

Lelu [443]

Answer:

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8 0

3 years ago

Write equations to show whether the solubility of either of the following is affected by pH: (b) Hg₂(CN)₂.

nata0808 [166]

Increasing pH will increase the solubility of the Hg2(CN)2 by shifting equilibrium to right side.

What is the meaning of OH in chemistry?

The chemical group, ion, or radical OH that consists of one atom of hydrogen and one of oxygen and is neutral or negatively charged.

Hg2(CN)2 + 2OH- ----> 2HgO(s) + 2HCN

adding OH- to the mercury(l) cyanide will cause the formation of the solid HgO.

therefore increasing pH will increase the solubility of the Hg2(CN)2 by shifting equilibrium to right side.

Learn more about OH

brainly.com/question/2911201

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3 0

2 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

What evidence supports a conservation law?

Murljashka [212]

Answer:

A- Energy from sunlight becomes glucose during photosynthesis

Explanation:

The law of conservational photosynthetic energy states that atoms cannot be created nor destroyed; the atoms can only be rearranged into new molecules and ions. This implies that the enzymes in photosynthesis are aided by energy to transform into glucose, which the plant needs.

5 0

2 years ago

Read 2 more answers

PLS HELP The average atomic mass of carbon is 12.01 amu. Based on the atomic

Masteriza [31]

Answer:

B. There is a very large percentage of C-12.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.

In such a way, the answer will be B. There is a very large percentage of C-12.

Regards!

7 0

3 years ago