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choli [55]
2 years ago
8

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

Chemistry
1 answer:
Sedaia [141]2 years ago
6 0

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

<u />

<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>

<em><u /></em>

The solute can be <em>crystalline or non-crystalline in nature.</em>

<em></em>

The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

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

a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>

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

Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

To learn more, visit the link below:

brainly.com/question/8482854

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Explain, in your own words, why different elements produce different colors of light when heated.
fgiga [73]
Different elements produce different colors of light when heated because the electrons in these elements have different permissible energy levels. When an element is heated, the electrons inside it become excited and move to an higher energy level from the ground state. When the electrons drop from this higher energy level, they typically emit energy quantum, the color of the light that is observed at this stage depends on difference that exist in the two energy levels.<span />
7 0
3 years ago
Read 2 more answers
The amount of matter in an object is called
marta [7]

Answer: Matter

Explanation:

Matter is anything that has volume and/or mass.

7 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
Drag and drop each phrase below the type of weathering it describes.
iren2701 [21]

Answer:

Mechanical weathering

A. does not change rock composition

C. abrasion

F. ice wedging

Chemical weathering

B. oxidabon

D. changes rock composition

E. acid rain

Explanation:

5 0
3 years ago
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the pressure of 5.0 L of gas increases from 1.50atm to 1240 mmgh. what is the final volume of the gas, assuming no change in mol
NikAS [45]

Answer:

V₂ = 4.7 L

Explanation:

Given data:

Initial volume = 5.0 L

Initial pressure = 1.50 atm

Final pressure = 1240 mmHg (1240/760 = 1.6 atm)

Final volume = ?

Solution:

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = 1.50 atm ×5.0 L/1.6 atm

V₂ = 7.5 atm. L /1.6 atm

V₂ = 4.7 L

6 0
3 years ago
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