Determine the mass of copper that has the same number of atoms as there are in 6.95 mg of potassium. Answer in units of mg.

Please someone solve this and tell me how you solve it

lianna [129]

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!

8 0

3 years ago

Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume

pishuonlain [190]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

mass = 0.508 g, Volume = 0.175 L

Temperature = (25 + 273) K = 298 K, P = 1 atm

As per the ideal gas law, PV = nRT.

where, n = no. of moles = $\frac{mass}{\text{molar mass}}$

Hence, putting all the given values into the ideal gas equation as follows.

PV = $\frac{mass}{\text{molar mass}} \times RT$

1 atm \times 0.175 L = $\frac{0.508 g}{\text{molar mass}} \times 0.0821 L atm/ K mol \times 298 K$

= 71.02 g

As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of $Cl_{2}$ is 70.8 g/mol or 71 g/mol (approx).

Thus, we can conclude that the gas is most likely chlorine.

4 0

3 years ago

The use of crucible

Norma-Jean [14]

a ceramic or metal container in which metals or other substances may be melted or subjected to very high temperatures.

"the crucible tipped and the mold filled with liquid metal"

a situation of severe trial, or in which different elements interact, leading to the creation of something new.

"their relationship was forged in the crucible of war"

3 0

3 years ago

A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25Â°.

ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

$pOH=-\log[OH^-]$

$1.00=-\log[OH^-]$

$[OH^-]=10^{-1.00} M=0.100 M$

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

$[KOH]=[OH^-]=[K^+]=0.100 M$

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

$Molarity=\frac{Moles}{Volume(L)}$

$0.100 M=\frac{n}{0.800 L}$

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

4 0

3 years ago

What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.4

sesenic [268]

Answer: The concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the $CaCl_2$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $AlCl_3$

We are given:

$n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL$

Putting all the values in above equation, we get

$M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M$

The concentration of $Cl^-$ ions in the resulting solution will be same as the molarity of solution which is 1.16 M.

Hence, the concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

6 0

3 years ago