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labwork [276]
3 years ago
10

How many grams of a 19.6% sugar solution contain 72.5 g of sugar?

Chemistry
1 answer:
anzhelika [568]3 years ago
4 0

Answer:

6.2g of sugar solution contains 72.5g

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Calculate number of moles in 10.6g of sodium carbonate
natulia [17]
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g 
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
7 0
3 years ago
What is the molarity of 4 qof NaCl (MM=58.45) in 3,800 mL of solution?
tiny-mole [99]

Answer:

.018 M

Explanation:

grams/MM=ans./volume(L) = M

4/58.45=ans./3.8=.018 M

8 0
3 years ago
A sample of an ionic compound containing iron and chlorine is analyzed and found to have a molar mass of 126.8 g/mol. what is th
julsineya [31]
Answer is: <span>the charge of the iron in this compound is +2.
Atomic mass of iron is 55,8 g/mol.
Atomic mass of chlorine is 35,5 g/mol.
If compound is FeCl, molar mass would be 55,8 </span>g/mol + 35,5 g/mol = 91,3 g/mo, that is not correct.
If compound is FeCl₂, malar mass of compound would be:
55,8 g/mol + 2·35,5 g/mol = 126,8 g/mol, that is correct.
Oxaidation number of chlorine is -1.
5 0
3 years ago
Read 2 more answers
I just need help plz!
Nesterboy [21]

Answer:

<u>Graphene</u> is a nanomaterial that is often used with other compounds to desalinate and decontaminate water.

The electrical behavior of semiconductor devices and electronics can be engineered using a process called <u>Doping</u>

Explanation:

4 0
3 years ago
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
3 years ago
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