What is the [H3O+]in a solution of pOH 0.253

how many moles of potassium would you need to prepare 1200 grams of 5.0% potassium sulfate (m/m) solution

Marianna [84]

The moles of potassium that you would need to prepare 1200 g of 5% potassium sulfate solution is 1.538 moles

calculation
calculate the mass potassium using the below formula

%M/M = mass of the solute(potassium)/mass of the solvent (potassium sulfate solution)

let the mass of potassium be represented by Y

then convert % into fraction = 5/100

5/100 = Y/1200
cross multiplication
100y = 6000
divide both side by 100

Y= 60 g

moles of potassium =mass/molar mass

= 60/39=1.538

7 0

2 years ago

An unknown element X has the following isotopes: ²⁵X (80.5% abundant) and ²⁷X (19.5% abundant). What is the average atomic mass

IgorC [24]

Answer:

25.39

Explanation:

Given parameters:

Abundance of X-25 = 80.5%

Abundance of X - 27 = 19.5%

Unknown:

Average atomic mass of X = ?

Solution:

The average atomic mass of X can be derived using the expression below:

Average atomic mass = (abundance x mass of X - 25) + (abundance x mass of X - 27)

Average atomic mass = (80.5% x 25) + (19.5% x 27) = 25.39

5 0

1 year ago

What is the correct name for the ionic compound Ca2P5

maks197457 [2]

Answer:

It is calcium phosphide.

Explanation:

Calcium phosphide is a salt like material that is composed of calcium and phosphorus in the ratio of 2:5. It has the appearance of red brown Crystalline salt.

It is use as rodenticide to kill rats.

It is also use in fire works.

7 0

2 years ago

Read 2 more answers

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

Answer: The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

A solution is prepared by mixing the 50 mL of 1 M NaH2PO4 with 50 mL of 1 M Na2HPO4. On the basis of the information, which of t

barxatty [35]

Answer:

A. PO43-

Explanation:

in the given is buffer . so H2PO4- and HPO42- both are present with equal concentration . Na+ is spectator ion it is also present in the concentration higher than the given species above .

but PO4-3 is not present . so it is lowest concentration

4 0

2 years ago