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3 years ago

In the compounds formed by li, na, k, rb, and cs with chlorine, how many atoms of cl are there per atom of the metal?

1 answer:

5 0

To determine how many chlorine atoms for the compound made from the metals given, we need to know the ionic forms of the metals above. Lithium when in ionic form would have a charge of +1. Sodium metal would have a +1 ionic charge. Potassium would have a charge of +1 as well. Rubidium would have a charge of +1. And Cesium would have +1 also. These metals are located at group 1A so they would all have a +1 charge in the ionic form. To balance this charge, -1 should be paired which would be the charge of a chloride ion. Therefore, for each metal, only 1 chlorine atom would bond.

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If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f

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Answer:

0.032 L or 32 mL

Explanation:

Use the dilution equation M1V1 = M2V2

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3 years ago

A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3

Vera_Pavlovna [14]

Answer:

$2.54\ \text{mg/L}$

Explanation:

C = Allowable concentration = 1.1 mg/L

$Q_1$ = Flow rate of river = $130\ \text{m}^/\text{s}$

$Q_2$ = Discharge from plant = $37\ \text{m}^3/\text{s}$

$C_1$ = Background concentration = 0.69 mg/L

$C_2$ = Maximum concentration that of the pollutant

The concentration of the mixture will be

$C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}$

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is $2.54\ \text{mg/L}$ .

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3 years ago