Two atoms that are isotopes of one another must have the same number of what?

What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

Therefore, concentration of potassium sulfate solution is 0.236 M.

4 0

3 years ago

¿Cuál de las siguientes configuraciones globales corresponde a un elemento químico que se comporta como metal? 1 punto [He]2s2 2

yuradex [85]

Answer:

[Ne]3s2

Explanation:

ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.

considerando la configuración electrónica más externa de cada una de las especies mostradas;

para la primera configuración, ns2 np6 corresponde a un gas noble.

para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.

para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.

para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6

5 0

3 years ago

And earthquake create a type of way that shakes the ground if a large earthquake occurs in Greece how can the waves be felt acro

Likurg_2 [28]

Earthquakes release seismic waves which can move along a sea.

3 0

4 years ago

Which is the limiting reagent in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2

NemiM [27]

Answer:

CuSO4

Explanation:

Na2S + CuSO4 → Na2SO4 + CuS

The reaction is balanced (same number of elements in each side)

To determine limiting reagent you need to know the moles you have of each.

Molar mass Na2S = 23 * 2 + 32 = 78

Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5

Na2S mole = 15.5 / 78 = 0.2

CuSO4 mole = 12.1/159.5 = 0.076

*Remember mole = mass / MM

With that information now you have to divide each moles by its respective stoichiometric coefficient

Na2S stoichiometric coefficient : 1

Na2S : 0.2 / 1 = 0.2

CuSO4 stoichiometric coefficient: 1

CuSO4: 0.076 / 1 = 0.076

The smaller number between them its the limiting reagent, CuSO4

8 0

3 years ago

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

3 years ago