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3 years ago

Plsssss help ii need help

Chemistry

1 answer:

gulaghasi [49]3 years ago

4 0

Answer:

wut u need help with

Explanation:

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4 years ago

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Natasha2012 [34]

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3 years ago

A total of 25.0 mL of 0.150 M potassium hydroxide (KOH) was required to neutralize 15.0 mL of sulfuric acid (H2SO4) of unknown c

lutik1710 [3]

1.50 M concentration. Hope you are happy

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4 years ago

You are given a stock solution of 500.0 mL of 1.00M magnesium chloride solution. Calculate the volume of the stock solution you

alexgriva [62]

Answer:

$50\; \rm mL$ of the stock solution would be required.

Explanation:

Assume that a solution of volume $V$ contains a solute with a concentration of $c$ . The quantity $n$ of that solute in this solution would be:

$n = c \cdot V$ .

For the solution that needs to be prepared, $c = 0.20\; \rm M = 0.20\; \rm mol \cdot L^{-1}$ . The volume of this solution is $V = 250.0\; \rm mL$ . Calculate the quantity of the solute (magnesium chloride) in the required solution:

$\begin{aligned}n &= c \cdot V \\ &= 0.20\; \rm mol \cdot L^{-1} \times 250.0\; \rm mL \\ &= 0.20\; \rm mol \cdot L^{-1} \times 0.2500\; \rm L \\ &= 0.050\; \rm mol\end{aligned}$ .

Rearrange the equation $n = c \cdot V$ to find an expression of volume $V$ , given the concentration $c$ and quantity $n$ of the solute:

$\displaystyle V= \frac{n}{c}$ .

Concentration of the solute in the stock solution: $c(\text{stock}) = 1.00\; \rm M = 1.00\; \rm mol \cdot L^{-1}$ .

Quantity of the solute required: $n = 0.050\; \rm mol$ .

Calculate the volume of the stock solution that would contain the required $n = 0.050\; \rm mol$ of the magnesium chloride solute:

$\begin{aligned}& V(\text{stock}) \\ &= \frac{n}{c(\text{stock})} \\ &= \frac{0.050\; \rm mol}{1.00\; \rm mol \cdot L^{-1}} \\ &= 0.050\; \rm L \\ &= 50\; \rm mL\end{aligned}$ .

3 0

3 years ago

An aqueous NaClNaCl solution is made using 111 gg of NaClNaCl diluted to a total solution volume of 1.20 LL .Calculate the molal

Aliun [14]

Answer:

The answer to your question is m = 1.6

Explanation:

Data

mass of NaCl = 111 g

volume of water = 1.2 L

Molality = ?

Formula

Molality = number of moles / volume (L)

Process

1.- Calculate the moles of NaCl

Molecular mass = 23 + 35.5 = 58.5 g

1 mol of ------------------- 58.5 g

x -------------------- 111 g

x = (111 x 1) / 58.5

x = 1.897 moles of NaCl

2.- Calculate the mass of water

density of water = 997 kg/m³

mass = volume x density

mass = 0.0012 x 997

mass = 1.19 kg

3.- Calculate molality

m = 1.897/1.19

m = 1.6

7 0

3 years ago