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zepelin [54]
3 years ago
15

9. How many grams of potassium sulfate are needed to make 250 mL of a 0.150 M

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

6.53g of K₂SO₄

Explanation:

Formula of the compound is K₂SO₄

Given parameters:

Volume of K₂SO₄ = 250mL = 250 x 10⁻³L

= 0.25L

Concentration of K₂SO₄ = 0.15M or 0. 15mol/L

Unknown:

Mass of K₂SO₄ =?

Methods:

We use the mole concept to solve this kind of problem.

>>First, we find the number of moles using the expression below:

Number of moles= concentration x volume

Solving for number of moles:

Number of moles = 0.25 x 01.5

= 0.0375mole

>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:

Mass(g) = number of moles x molar mass

Solving:

To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.

For:

K = 39g

S = 32g

O = 16g

Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)

= 78 +32 + 64

= 174g/mol

Using the expression:

Mass(g) = number of moles x molar mass

Mass of K₂SO₄ = 0.0375 x 174 = 6.53g

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46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride
slega [8]

Answer:

m_{HgCl_2}=42.7gHgCl_2

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2

Best regards.

3 0
3 years ago
The atomic particle with a charge of -1.6 x 10-19 C is
kobusy [5.1K]

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7 0
3 years ago
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5 0
2 years ago
What is the approximate percent by mass of potassium in KHCO3?
lozanna [386]

Answer:

The mass percent of potassium is 39%

Option C is correct

Explanation:

Step 1: Data given

Atomic mass of K = 39.10 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molar mass of KHCO3

Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0

Molar mass KHCO3 = 100.12 g/mol

Step 3: Calculate mass percent of potassium (K)

%K = (atomic mass of K / molar mass of KHCO3) * 100%

%K = (39.10 / 100.12) * 100%

%K = 39.05 %

The mass percent of potassium is 39%

Option C is correct

8 0
3 years ago
How is a coefficient used to balance an equation
ludmilkaskok [199]

Let's start to understand this question by a simple combustion reaction involving oxidation of Ethane in the presence of Oxygen. When Ethane is burned in the presence of Oxygen it produces Carbon Dioxide and Water respectively. Therefore, the equation is as,

                                C₂H₆  +  O₂    →    CO₂  +  H₂O

Above reaction shows the reaction and the equation is unbalanced. Balancing chemical equation is important because according to law of conservation of mass, mass can neither be created nor destroyed. Hence, we should balance the number of elements on both side.

                                       LHS                      RHS

Carbon Atoms                  2                            1

Hydrogen Atoms              6                           2

Oxygen Atoms                  2                           3

It means this equation is not obeying the law. Now, how to balance? One way is as follow,

                                C₂H₆  +  O₃    →    C₂O₂  +  H₆O

                                       LHS                      RHS

Carbon Atoms                  2                            2

Hydrogen Atoms              6                           6

Oxygen Atoms                  3                           3

We have balanced the equation by changing the subscripts. But, we have messed up the chemical composition of compounds and molecules like Oxygen is converted into Ozone.

Therefore, we will change the coefficients (moles) to balance the equation as,

                                C₂H₆  +  7/2 O₂    →    2 CO₂  +  3 H₂O

                                       LHS                      RHS

Carbon Atoms                  2                            2

Hydrogen Atoms              6                           6

Oxygen Atoms                  7                           7

Now, by changing the coefficients we have balanced the equation without disturbing the chemical composition of compounds and molecules.

3 0
3 years ago
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