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Nady [450]
3 years ago
10

What will happen when we add 5 grams of sugar to 100 grams of water? Plz tell

Chemistry
2 answers:
Anarel [89]3 years ago
8 0
<h3>♫ :::::::::::::::::::::::::::::: // Hello There ! //  :::::::::::::::::::::::::::::: ♫</h3>

➷  It will most likely dissolve in the water. This is due to the sugar being a soluble substance.

Any further queries, let me know.

<h3><u>❄️</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

Effectus [21]3 years ago
8 0

Answer:

mass of solute = 5g

mass of solution = 5g +100g =105g

mass/mass percentage x 100

5/105 x 100 =25 x 100 = 2500

Explanation:

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One isotope of oxygen differs from another isotope of oxygen in *
Reptile [31]

Each isotope of Oxygen has a different number of neutrons

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Atomic mass is the average atomic mass of all its isotopes

Isotopes are atoms has the same number of protons but has a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Some of the isotopes of oxygen are:

\tt _8^{16}O,_8^{17}O,_8^{18}O

Each isotope has 8 protons and 8 electrons but has a different number of neutrons

For O-16: number of neutrons = 16-8 = 8

For O-17: number of neutrons = 17-8 = 9

For O-18: number of neutrons = 18-8 = 10

4 0
3 years ago
How many moles are in 155 grams of aluminum
Aliun [14]

Answer:

wads

Explanation:

wadsdw

8 0
3 years ago
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
3 years ago
What is the minimum volume of 5.50 M HCl necessary to neutralize completely the hydroxide in 718.0 mL of 0.183 M NaOH
Deffense [45]

The volume of HCl required is 23.89 mL

Calculation of volume:

The reaction:

HCl + NaOH \rightarrow NaCl + H_2O

As HCl and NaOH react in 1 : 1 ratio.

Volume of NaOH= 718 mL

Concentration= 0.183M

Volume of HCl= ?

Concentration= 5.50M

Using the dilution formula:

M_1\times V_1(NaOH)= M_2\times V_2(HCl)

0.183\times 718= 5.50 \times V_2\\V_2=\frac{131.394}{5.50} \\V_2 = 23.89 \,mL

Therefore,

Volume of HCl required will be 23.89 mL.

Learn more about neutralization reaction here,

brainly.com/question/1822651

#SPJ4

6 0
2 years ago
Please help asap
Helen [10]
C. carbon
sodium is an
oxygen is o
cl is chlorine
4 0
3 years ago
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