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3 years ago

Multiple choice. Please help.

Chemistry

1 answer:

Ludmilka [50]3 years ago

7 0

Answer:

7. B

8. B

9. A

10. A

11. Increases

Again..I hope it helps.✌

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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 $565.414\times 10^{-6}g/L$ .

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1) All atoms are _____.

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A patient receives 3.3 L of glucose solution intravenously (IV). If 100. mL of the solution

artcher [175]

Answer:

660kcal

Explanation:

The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.

We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:

3.3L * (1000mL / L) * (5g/100mL)= 165 g.

If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.

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4 years ago

Which object can electrically polarized

kobusy [5.1K]

Answer:Static electricity works because objects which are otherwise "neutral" (in other words, objects with no net charge) can be polarized. An electric field, like one caused by a nearby charged object, can cause the charges inside of a neutral object — the protons and electrons — to move around a tiny bit.

Explanation:

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