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netineya [11]
3 years ago
11

What mass of potassium bromide (in grams) do you need to make 250.0 mL of a 1.50 M potassium bromide solution?

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
8 0

Answer:

44.63g

Explanation:

First, let us calculate the number of mole of KBr in 1.50M KBr solution.

This is illustrated below:

Data obtained from the question include:

Volume of solution = 250mL = 250/1000 = 0.25L

Molarity of solution = 1.50M

Mole of solute (KBr) =.?

Molarity is simply mole of solute per unit litre of solution

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of solute (KBr) = 1.50 x 0.25

Mole of solute (KBr) = 0.375 mole

Now, we calculate the mass of KBr needed to make the solution as follow:

Molar Mass of KBr = 39 + 80 = 119g/mol

Mole of KBr = 0.375 mole

Mass of KBr =?

Mass = number of mole x molar Mass

Mass of KBr = 0.375 x 119

Mass of KBr = 44.63g

Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution

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svetoff [14.1K]

Answer:

period 3 and group 3

Explanation:

I'm saying group 3 because that is how I learnt it at school, but if you count it then it's in group 13.

8 0
3 years ago
Which statement describes the reactions in an electrochemical cell
34kurt

Answer & explanation:

Summary on electrochemical cells and redox reactions:

Electrochemical cells (or batteries) can be defined as devices capable of transforming chemical energy into electrical energy through spontaneous reactions of redox, in which electron transfer occurs.

Redox it is a chemical reaction in which there is the occurrence of oxidation and reduction of atoms of substances (chemical species) present in the process.

<u>Oxidation</u> is the loss of electrons by an atom of a chemical species, while <u>reduction</u> is the gain of electrons by an atom of a chemical species.

Thus, during an oxirreduction reaction, electrons move from the species that loses them towards the species that will receive them. This "movement" results in the formation of an electric current (or electrical energy) as occurs with batteries, for example.

6 0
4 years ago
Calculate the volume (liters) of solution that can be prepared from 180. grams of K2Cr2O7 for a 0.200M solution.
Sladkaya [172]

Answer:

The volume of the solution is 0.305 liters.

Explanation:

Molar mass is the amount of mass that a substance contains in one mole. The molar mass of K₂Cr₂O₇ is 294 g / mole. Then you can apply the following rule of three: if by definition of molar mass 294 grams of the compound are contained in 1 mole, 180 grams are contained in how many moles?

moles=\frac{180 grams*1mole}{294 grams}

moles= 0.61

Molarity is a measure of the concentration of a substance that is defined as the number of moles contained in a certain volume. So, the molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}.

In this case:

  • molarity= 2 M
  • number of moles of solute= 0.61 moles
  • volume= ?

Replacing in the definition of molarity:

2 M=\frac{0.61 moles}{volume}

Solving:

volume=\frac{0.61 moles}{2 M}

volume= 0.305 liters

<u><em>The volume of the solution is 0.305 liters.</em></u>

8 0
3 years ago
Suppose a salt and a glucose solution are separated by a membrane that is permeable to water but not to the solutes. the nacl so
stira [4]
1)

<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

</span>V(H2O) = 250ml = 250g = 0.25 kg<span>

</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole

</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation

Since NaCl dissociates into 2 ions in solution:
<span>                                        
</span>milliosmolality of NaOH = 132 x 2 = 264  osmol<span>es/kg
</span>
2)

m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

V(H2O) = 250ml = 250g = 0.25 kg

molality of glucose:

n(gl)=m/M=9/180= 0.05 mole

molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg

milimolality of glucose = 0.132/0,001 = 200 mmole/kg

milliosmolality of glucose = milimolality x N of ions formed in dissociation

Since glucose does not dissociate, milimolality and milliosmolality are same:
                                        
milliosmolality of glucose = 200 osmoles/kg

3)

The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.

In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.

3 0
4 years ago
A pure compound is found to be 40.0% carbon by mass, 6.73% hydrogen by mass, and 53.3% oxygen by mass. determine the empirical f
ratelena [41]
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O

With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
3 0
3 years ago
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