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Ierofanga [76]
3 years ago
15

19. When sugar is dissolved in water

Chemistry
1 answer:
Mars2501 [29]3 years ago
6 0
C.
The solute is dissolved in the solvent
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If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial
jeka94

Answer:

the  initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

KSCN \rightarrow K^ + SCN^-

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5  m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the  initial concentration of SCN- in the mixture is 0.00588 M

5 0
3 years ago
Given 2Na + Cl2=2NaCl, what is the excess reactant? What is the limiting reactant?
QveST [7]

Answer: The limiting reactant is Na

Explanation:

5 0
3 years ago
Question 2
Pavel [41]
Tearing paper, Physical Change
6 0
3 years ago
The reducing agent in a material is the one that A. gains electrons. B. increases its atomic number. C. is reduced. D. is oxidiz
Bas_tet [7]
D. Is Oxidized Hope I helped.
4 0
3 years ago
Read 2 more answers
An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having
dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass  2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007  (1-x) = 1.008  

2.014x + 1.007  - 1.007 x = 1.008  

2.014x - 1.007x  =  1.008  -  1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass  2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0
3 years ago
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