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hjlf
3 years ago
5

Increasing the ? of a solvent increases the solubility of the solute

Chemistry
2 answers:
mart [117]3 years ago
8 0

The answer is temperature

Gelneren [198K]3 years ago
7 0
There are a lot of ways to increase the solubility of the solute. <span>Increasing the temperature, mixing time and surface area of a solvent increases the solubility of the solute</span>
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What type of radiation is likely to occur when the ratio of protons to neutrons is below the band of stability? a. alpha decay b
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Beta radiation / decay would likely occur when the ratio of protons to neutrons is below the band of stability.
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3 years ago
Which of the following is an example of genetic engineering
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Answer:

B:  Inserting a gene from a flounder into salmon DNA to produce antifreeze proteins.

Explanation:

Hope this helps.

5 0
3 years ago
Read 2 more answers
What is SO2 shape name?
muminat

Answer:Molecular Formula SO2

Hybridization Type sp2

Bond Angle 119o

Geometry V-Shaped or Bent

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5 0
3 years ago
The volume of air in a person’s lungs is 615 mL at a pressure of 760. mmHg. Inhalation occurs as the pressure in the lungs drops
chubhunter [2.5K]

<u>Answer:</u> The final volume of lungs is 621.5 mL

<u>Explanation:</u>

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=760mmHg\\V_1=615mL\\P_2=752mmHg\\V_2=?mL

Putting values in above equation, we get:

760mmHg\times 615mL=752mmHg\times V_2\\\\V_2=\frac{760\times 615}{752}=621.5mL

Hence, the final volume of lungs is 621.5 mL

8 0
3 years ago
HELP. NO FAKE ANSWERS. I WILL REPORT. I AM CONFUSED AND NEED HELP. FILL IN THE NOT FILLED BOXES POR FAVOR.
SCORPION-xisa [38]
Question #1
Potasium hydroxide (known)
 volume used is 25 ml 
Molarity (concentration) = 0.150 M
Moles of KOH used 
           0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4) 
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l) 
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
      0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid)  concentration
    0.001875 moles × 1000/15  
        = 0.125 M

Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
      0.25 × 30/1000
        =  0.0075 moles 
Sodium Hydroxide (base)
Volume used 20 ml 
Concentration (unknown)
The equation for the reaction is 
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr   is 1 : 1
Therefore, moles of NaOH used;
                 = 0.0075 moles
NaOH concentration will be 
       = 0.0075 moles × 1000/20
       = 0.375 M

7 0
3 years ago
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