Please someone solve this and tell me how you solve it

Chemistry

1 answer:

lianna [129]3 years ago

8 0

Answer:

Supersaturated.

Explanation:

Hello there!

In this case, according to this solubility chart, we infer that for NH3, the solubility starts at 90 grams of NH3 that are soluble in 100 g of water at 0 °C and ends in about 8 g in 100 g of water at 100 °C for a saturated solution.

However, since we are asked for the solubility of NH3 at 20 °C, we can see that, according to the table and the curve for NH3, about 52 g of NH3 are soluble in 100 g of water; thus, for the given 60 g of NH3, we will say that 8 grams will remain undissolved, and therefore, this solution will be supersaturated.

Regards!

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9. To become like a noble gas the element P will have a charge of

MrRa [10]

Answer:

to become a noble gas element P will have 2 electrons in it's outer most energy level if it has one energy level

and eight in the last energy level if more than one

6 0

3 years ago

A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?

Katen [24]

V ( HCl ) = 16.4 mL / 1000 => 0.0164 L

M( HCl) = ?

V( KOH) = 12.7 mL / 1000 => 0.0127 L

M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

HCl + KOH = H2O + KCl

1 mole HCl ------ 1 mole KOH
? mole HCl--------0.007874 moles KOH

moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / 0.0164

= 0.480 M

Answer (2)

hope this helps!

4 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$