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3 years ago

Covert 0.2598kl to l

1 answer:

7 0

Kl in this problem means kilo liter. 1 kilo liter is equivalent to 1000 liter and equal to one cubic meter.

But in this problem, you are asking to convert this to liter.

So:

0.2598 kl = 1000 liter

--------------

1 kilo liter

So multiply, we get 259.8 liters or 260 liters.

You might be interested in

Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.

tamaranim1 [39]

Answer:

1500kg/m^3

Explanation:

Formula:

d=m/v

Given:

m=1.5kg

v=1000cm^3

(The side length of a cube is always equal to the others)

Required:

d=?

Solution:

d=m/v

d=1.5kg/1000cm^3

d=1.5kg/0.001m^3

d=1500kg/m^3

Hope this helps ;) ❤❤❤

3 0

3 years ago

A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0

3 years ago

Why was alchemy so confusing?

galina1969 [7]

Search Results
Featured snippet from the web
Alchemy is extremely complicated. It is based on the practical skills of early metal workers and craftsmen, on Greek philosophy, and on Eastern mystic cults that sprang up in the first centuries after Christ and influenced so much of magic and occult thought.

8 0

2 years ago

A solution that is 0.20 m in hcho2 and 0.15 m in nacho2 find ph

Mashutka [201]

We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH

The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])

We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87

So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.

5 0

3 years ago

For each phenotype, write the possible genotypes.

Citrus2011 [14]

Answer:

Tall, Res, Dom

short, RES, RES

3 0

3 years ago