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valentina_108 [34]
3 years ago
6

How can you read a determine the masses

Chemistry
1 answer:
krok68 [10]3 years ago
4 0
I need you to explain the question
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Jonathan claims that since only humans can read, it must be an inherited trait, passed from human parents to their children. Is
Crank
Jonathan is not correct because it has to be a trait that is learned.

So in that case Jonathan has to say that this is not an inherited trait it is learned by most people.
5 0
3 years ago
A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3p
timurjin [86]

Answer:

55.75g

Explanation:

From

m/M = CV

Where

m= required mass of solute

M= molar mass of solute

C= concentration of solution

V= volume of solution=675ml

Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1

Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles

Since 1 mole of Na3PO4 contains 3 moles of Na+

It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles

mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g

3 0
3 years ago
How can indicator be used to determine if a solution is an acid or base
dalvyx [7]

Answer:

To tell if something is an acid or a base, you can use a chemical called an indicator. An indicator changes color when it encounters an acid or base. There are many different types of indicators, some that are liquids and others that are concentrated on little strips of "litmus" paper.

Explanation:

8 0
3 years ago
2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu; 83.79% of 52Cr, with an
shutvik [7]

Answer:

Average atomic mass  = 51.9963 amu

Explanation:

Given data:

Abundance of Cr⁵⁰ with atomic mass= 4.34% ,  49.9460 amu

Abundance of Cr⁵² with atomic mass = 83.79%,  51.9405 amu

Abundance of Cr⁵³ with atomic mass =9.50%,  52.9407 amu

Abundance of Cr⁵⁴ with atomic mass  = 2.37%,   53.9389 amu

Average atomic mass = 51.9963 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n)  / 100

Average atomic mass  = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100

Average atomic mass =  216.7656 + 4352.0945 + 502.9367 +127.8352 / 100

Average atomic mass  = 5199.632 / 100

Average atomic mass  = 51.9963 amu

4 0
3 years ago
A 50.00 L sample of gas collected in the upper atmosphere at a pressure of 18.30 torr is compressed into a volume of 0.1500 L. W
andreev551 [17]

Answer:

the ans is in the picture with the steps  

(hope it helps can i plz have brainlist :D hehe)

Explanation:

5 0
3 years ago
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