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2 years ago

Graphite and iodine are non metal but they shine why secondry school

Chemistry

1 answer:

Alex_Xolod [135]2 years ago

8 0

It is their properties. it is made up naturally due to the ions

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How do ions form bonds, and describe the structure of the resulting compound.

Artemon [7]

Answer:

Electrons are transferred between atoms together in the ionic compound. The ions are arranged in a regular repeating pattern in an ionic crystal

Explanation:

6 0

2 years ago

Which of the following is true about elements?

kondor19780726 [428]

Answer:

The first one

Explanation:

Elements are as simple as it gets

6 0

2 years ago

What is the molarity of 2.00 L of a solution that contains 14.6 g NaCl?

White raven [17]

Answer: The molarity of the solution is 0.125 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

moles of $NaCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{14.6g}{58.5g/mol}=0.250mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.250mol}{2.00L}=0.125M$

Therefore, the molarity of the solution is 0.125 M

8 0

2 years ago

One mole of H atom is equal to

valentina_108 [34]

16 grams I think it might be it

6 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

2 years ago