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3 years ago

The graph shows a solubility curve of a soluble substance.

Chemistry

1 answer:

DochEvi [55]3 years ago

5 0

Answer:

Looks to be almost at 30

Explanation:

Based on the graph

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Another student whose name is geraldine brought 2 litro packs of orange juice and 6 liters of bottled water. the ideal mixture t

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2 years ago

When testing for hydrogen gas in the laboratory a glowing splint is inserted into the gas which results in a pop sound. What is

Advocard [28]

C. Exothermic reaction I guess

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2 years ago

Please help ASAP

marta [7]

Answer:

0.1

Explanation:

We must first put down the equation of the reaction in order to guide our solution of the question.

2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)

Now from the question, the following were given;

Concentration of acid CA= ??????

Concentration of base CB= 0.299M

Volume of acid VA= 17.8ml

Volume of base VB= 24.7ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

SUBSTITUTING VALUES;

CA= 0.299 × 24.7 ×2 / 17.8×1

CA= 0.8298 M

But;

pH= -log[H^+]

[H^+] = 0.8298 M

pH= -log[0.8298 M]

pH= 0.1

5 0

3 years ago

The molecule that carries an amino acid to the ribosome for incorporation into a protein is ________.

alukav5142 [94]

TRNAs (transfer RNAs) carry amino acids to the ribosome. They act as "bridges," matching a codon in an mRNA with the amino acid it codes for.

8 0

3 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

3 years ago