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2 years ago

What is the formula and formula mass of potassium sulphate?

Chemistry

2 answers:

Norma-Jean [14]2 years ago

7 0

Pretty sure the formula is K2SO4.

olchik [2.2K]2 years ago

7 0

Formula: K2SO4

Molar mass: 174.259 g/mol

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If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

Natalia dipped two feathers in oil. Then she dipped one feather in cold water and the other feather in hot water. She swirled bo

Alika [10]

Answer:

The hot water was better for removing the oil.

Explanation:

You can see that because the mass went down more with the hot water. So, that means that more oil was taken out of the feather with hot water.

8 0

2 years ago

Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles

Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃ + O₂ ⟶ N₂ + H₂O

The balanced reaction:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0

3 years ago

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

<u>Answer:</u> The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago

Two samples of a compound containing elements a and b are decomposed. the first sample produces 15 g of a and 35 g of

Alborosie

According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.

The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.

Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases

\frac{a}{b}=\frac{15}{35}=\frac{10}{y}

rearranging,

y=\frac{10\times 35}{15}=23.3

Thus, mass of b produced will be 23.3 g.

3 0

3 years ago