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Yuri [45]
3 years ago
15

Element X reacts with chlorine to form an ionic compound that has the formula XCl₂. To which group on the Periodic Table could e

lement X belong?
Chemistry
1 answer:
marishachu [46]3 years ago
4 0

Answer:

<u>Group 2 is the correct answer.</u>

Explanation:

<u>Chlorine has a charge -1. It means that X is +2 so that the formula is XCl2. </u>

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A sports car and a fully loaded semi truck pull side by side at a red light. Using your knowledge of acceleration, predict which
Len [333]
According to Newtons second law of motion:

a =  \frac{F}{m}

This shows, acceleration and mass are inversely proportional. So, larger the mass, less will be the acceleration produced and vice versa.

In case of a sports car and a fully loaded truck, the mass of truck is a lot more than the sports car. As a result the sports car will accelerate faster when the traffic light turns green.
7 0
3 years ago
The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.
Vikki [24]

Answer :  The solubility of this compound in g/L is 565.414\times 10^{-6}g/L.

Solution : Given,

K_{sp}=2.42\times 10^{-11}

Molar mass of MnCO_3 = 114.945g/mole

The balanced equilibrium reaction is,

                      MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3

At equilibrium                         s       s

The expression for solubility constant is,

K_{sp}=[Mn^{2+}][CO^{2-}_3]

Now put the given values in this expression, we get

2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L

Therefore, the solubility of this compound in g/L is 565.414\times 10^{-6}g/L.


8 0
2 years ago
A student dissolves 0.0688 mol of sodium hydroxide in water to make a 0.250 L solution. What is the molarity of the solution?
den301095 [7]
Number of moles = 0.0688 moles of NaoH

volume = 0.250 L

Molarity = moles of solute / volume ( L )

M = 0.0688 / 0.250

M = 0.28 M

Answer B
7 0
3 years ago
The concentration of hydrogen peroxide solution can be determined by
max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0
3 years ago
What explanation can you give for why the sodium-potassium pump does not run out of ions to move in or out of the cell
rewona [7]

The sodium-potassium pump does not run out of ions since ion exchange is essential for the action potential to take place and to maintain homeostasis.

The cell has variable concentrations of different substances compared to the environment that surrounds it, with significant differences with sodium and potassium.

  • The main function of the sodium-potassium pump is to maintain homeostasis of the intracellular medium, controlling the concentrations of these two ions.

  • In order to carry out the adequate exchange of sodium and potassium ions in the extra and intracellular medium, the cells need an active transport process that is carried out thanks to the sodium potassium pump.

  • This process is needed for the maintenance and functioning of cells, and it is essential for the action potential to be executed, necessary for the transmission of electrical impulses from neuron to neuron.

Therefore, we can conclude that the sodium potassium pump produces an exchange of potassium ions for sodium ions which keeps the cellular system functioning properly.

Learn more here: brainly.com/question/24336764

6 0
2 years ago
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