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makvit [3.9K]
3 years ago
9

Question 2 of 5

Chemistry
2 answers:
alex41 [277]3 years ago
7 0
Precision relates to how close the answers are to each other, so I’d think it would be D because of the limited range between data points.
belka [17]3 years ago
6 0
A Group c 32.1 g 35.0 g 24.0 g
You might be interested in
You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution
vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

0.173\%=\frac{779 mg}{m}\times 100

m=\frac{779 mg}{0.173}\times 100= 450,289 mg

1 mg = 0.001 g

m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g

450. g of 0.173 % KCN solution contains 779 mg of KCN.

6 0
3 years ago
Position (m)
Alex777 [14]

Option C. The object is returning to the start at a constant speed.

<h3>Data points of the Position vs Time graph</h3>

The following data points will be used to determine the motion of the object.

<u>Position               Time</u>

12                          4

10                          6

2                            8

0                           10

From the data above, the position of the object is decreasing towards zero or start point.

Thus, the object is returning to the start at a constant speed.

Learn more about position here: brainly.com/question/2364404

#SPJ1

8 0
2 years ago
Which of the following elements is the most electronegative?
Veseljchak [2.6K]
Chlorine is the most electronegative
5 0
3 years ago
Read 2 more answers
The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is
iogann1982 [59]

Answer:

2.28 × 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^- + H2O ⇌ HCN + OH^-

                    Ka = 4.9 × 10^-10

               KaKb = Kw

4.9 × 10^-10 Kb = 1.00 × 10^-14

                   Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5

Now, we can set up an ICE table

                     CN^- + H2O ⇌ HCN + OH^-

I/(mol/L)      0.255                     0         0

C/(mol/L)       -x                        +x        +x

E/(mol/L)  0.255 - x                   x         x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

    x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

        x = sqrt(5.20 × 10^-6)    = 2.28 × 10^-3

[OH^-] = x mol/L                     = 2.28 × 10^-3 mol/L

5 0
3 years ago
If the half-life of a radioactive element is 4 days, how long will it take for three- fourths of a sample of the element to deca
Jlenok [28]

Answer:

\boxed{\text{8 da}}

Explanation:

The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

The half-life of the element is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:

\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}

\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}

3 0
3 years ago
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