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3241004551 [841]
3 years ago
12

If a “universal solvent” could dissolve anything, what problems might you imagine that would cause in everyday life?

Chemistry
1 answer:
algol133 years ago
3 0
People call water a 'universal solvent' because it is capable of dissolving more<span> substances than any other liquid. I think</span> it could<span> can be a major problem if every substance was readily soluble by water or any solvent. If so, it would mean that there is nothing that could contain water if it was not completely saturated with another solute. All in all, t</span><span>he idea of a universal solvent would be just impossible to imagine.</span>
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Suppose you wanted to dissolve 109 g of sodium carbonate in enough water to make 6.00 L of solution. What is the molarity of thi
kumpel [21]
Molarity

a. Na2CO3
                Molar mass = 106 g/mol
    n= 109 g / (106 g/mol) = 1.028 mol

b. 6.00L

c. M=n/V
    M = 1.028 mol / 6.00 L
    M= 0.1713 mol / L

Answer: The molarity of the sodium carbonate solution is 0.1713 mol/L

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4 years ago
Which two substances are among the four major types of organic
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Answer:

B.Carbohydrates and D. water

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3 years ago
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vlada-n [284]

d.

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3 years ago
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The half-life of bismuth-210, 210bi, is 5 days. (a) if a sample has a mass of 184 mg, find the amount remaining after 15 days.
ASHA 777 [7]

Answer:- 23.0 mg

Solution:- Radioactive decay obeys first order kinetics and the first order kinetics equation is:

lnN=-kt+lnN_0

where, N_0 is the initial amount of radioactive substance and N is it's amount after time t. k is the decay constant.

From given information, Original amount, N_0 of the radioactive substance is 184 mg and we are asked to calculate the amount N after 15 days. It means, t = 15 days

Half life is given as 5 days. From the half life, we could calculate the decay constant k using the equation:

k=\frac{0.693}{t_1_/_2}

where, t_1_/_2 is the symbol for half life. let's plug in the value of half like to calculate k:

k=\frac{0.693}{5days}

k=0.1386day^-^1

Let's plug in the values in the first order kinetics equation and solve it for N:

lnN=-0.1386day^-^1(15days)+ln184mg

lnN=-2.079+5.215

lnN = 3.136

N=e^3^.^1^3^6

N = 23.0 mg

So, 23.0 mg of Bi-210 would be remaining after 15 days.

3 0
4 years ago
2.98 moles of H2 at 35°C and 2.3 atm are in a 32.8 L container. How many moles of H2 are in a 45.3 L container under the same co
zalisa [80]

Answer:

4.12 moles

Explanation:

We can solve this problem with the Ideal Gases Law.

P . V = n . R . T

In our first case we have:

P = 2.3 atm

V = 32.8 L

n = 2.98 moles

T → 35°C + 273 =  308K

Let's replace data for the second case:

2.3 atm . 45.3L = n . 0.082 . 308K

n = (2.3 atm . 45.3L) / (0.082 L.atm/mol.K . 308K)

n = 4.12 moles

6 0
3 years ago
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