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1 year ago

Ima 5% glucose solution, how many grams of glucose would be present per 100mL

Chemistry

1 answer:

Sedaia [141]1 year ago

6 0

Percentage Weight-in-volume is defined as the number of grams of a solute in a 100 ml (milliliters) solution.

Percentage Weight-in-volume can tell us about the degree of concentration of a given solution.

The solute can be crystalline or non-crystalline in nature.

The number of grams of glucose present in a 5% glucose solution is 5 grams.

This question is based on a Percentage Weight-in-volume. The formula states that:

a% of a glucose solution = a grams of glucose in a 100 mL solution

Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution

Therefore, the number of grams of glucose present in a 5% glucose solution is 5 grams.

To learn more, visit the link below:

brainly.com/question/8482854

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Two proteins interact to form a multimeric complex. When one of the proteins is mutated, there is a substantial loss of function

ololo11 [35]

Answer:

<h2>4. dominant negative</h2>

Explanation:

Mutation is the process in which sudden changes take place within the sequence of amino acids that causes different type of problems. On the basis of nature and conditions mutation can be classified as dominant negative, neomorphic and some others types. Dominant mutation is also called as antimorphic mutation and changes the functions of the molecules that are proteins.

6 0

3 years ago

15.5 g of an unknown metal at 165.0°C is dropped into 150.0mL of H2O at 23.0°C in a coffee cup calorimeter. The metal and H2O re

tino4ka555 [31]

Answer:

Specific heat capacity of metal is 2.09 j/g.°C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of metal = 15.5 g

Initial temperature = 165.0°C

Initial temperature of water = 23.0°C

Final temperature = 30.0°C

Specific heat capacity of metal = ?

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 150.0 mL or 150.0 g

Solution:

Formula:

- Qm = +Qw

Now we will put the values in formula.

-15.5 g × c × [ 30.0°C - 165.0°C] = 150 g × 4.184 J/g°C × [ 30.0°C - 23.0°C]

15.5 g × c × 135°C = 4393.2 j

2092.5 g.°C × c = 4393.2 j

c = 4393.2 j/2092.5 g.°C

c = 2.09 j/g.°C

4 0

2 years ago

Write an equilibrium equation that shows bisulfate acting as a weak acid in water.

NemiM [27]

Answer:

Explanation:

Bisulphate ion is a weak acid as it can form hydronium ion in water .

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

The equilibrium constant of this reaction is very small , hence bisulphate ion is very weak acid.

4 0

3 years ago

2. What mass of water absorbs 6700 J of heat to raise the temperature from 283K to 318K?

Alexxandr [17]

Answer:

Q = mcT ...you can either substitute the molar heat capacity of water in the place of c or the specific heat capacity of water.

Explanation:

7 0

2 years ago

Write the net ionic equation for the reaction between hydrochloric acid and potassium cyanide. Do not include states such as (aq

zhannawk [14.2K]

Answer: $H^++CN^-\rightarrow HCN$ , product favoured

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we do not include the spectator ions in the equations.

When hydrochloric acid react with potassium cyanide, then it gives potassium chloride and hydrocyanic acid as products.

The complete ionic equation will be:

$H^++Cl^-+K^++CN^-\rightarrow K^++Cl^-+HCN$

The net ionic equation will not contain spectator ions which are $K^+$ and $Cl^-$ :

$H^++CN^-\rightarrow HCN$

The reaction is product favoured.

5 0

3 years ago