Nonmetals are __

What is the concentration (m) of sodium ions in 4.57 l of a 2.35 m na3p solution?

Leokris [45]

Answer: 1.03 x 10^25 ions Na

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3 years ago

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A

NikAS [45]

Answer:

After you have waived your rights, the confession must still be a voluntary confession. Whether a confession is voluntary or not is based on the “totality of the circumstances”. The Courts have held that a police officer simply lying does not result in an “involuntary confession”.

Explanation:

7 0

2 years ago

Can someone please help me with this chemistry hw plzzzzz I really need help

Viktor [21]

Hey I can't see the questions properly

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3 years ago

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If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4

attashe74 [19]

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$ .

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of $SO_4^{2-}$ , it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

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3 years ago

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A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH4NO3 and 4.42 g of (NH4)3PO4 in enough

motikmotik

Answer:

0.00725 M

Explanation:

Considering for $NH_4NO_3$

Mass = 5.66 g

Molar mass of $NH_4NO_3$ = 80.043 g/mol

Moles = Mass taken / Molar mass

So,

Moles = 5.66 / 80.043 moles = 0.0707 moles

$NH_4NO_3$ will dissociate as:

$NH_4NO_3\rightarrow NH_4^++NO_3^-$

Thus 1 mole of $NH_4NO_3$ yields 1 mole of ammonium ions. So,

Ammonium ions furnished by $NH_4NO_3$ = 1 × 0.0707 moles = 0.0707 moles

Considering for $(NH_4)_3PO_4$

Mass = 4.42 g

Molar mass of $(NH_4)_3PO_4$ = 149.09 g/mol

Moles = Mass taken / Molar mass

So,

Moles = 4.42 / 149.09 moles = 0.0296 moles

$(NH_4)_3PO_4$ will dissociate as:

$(NH_4)_3PO_4\rightarrow 3NH_4^++PO_4^{3-}$

Thus 1 mole of $NH_4NO_3$ yields 3 moles of ammonium ions. So,

Ammonium ions furnished by $(NH_4)_3PO_4$ = 3 × 0.0296 moles = 0.0888 moles

Total moles of the ammonium ions = 0.0707 + 0.0888 moles = 0.1595 moles

Given that:

Volume = 22.0 L

So, Molarity of the $NH_4^+$ is:

Molarity = Moles / Volume = 0.1595 / 22 M = 0.00725 M

8 0

3 years ago

Nonmetals are ___.