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IgorC [24]
3 years ago
5

What tool do you use to measure mass?

Chemistry
1 answer:
pshichka [43]3 years ago
7 0

Answer: Balance

Explanation:  Mass is the amount of matter contained in a body.

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Atoms of _______ gain electrons to fill their outer electron shells and become ________ ions.
Bezzdna [24]
Nonmetals and negative. Answer is c
6 0
3 years ago
Which pure substance is made of only one kind of atom?
aleksley [76]

Answer:

the answer is D i believe. i am not shure but im like 70% sure

7 0
3 years ago
If a balloon initially has a volume of 4.0 liters and a temperature of 10 degrees Celsius, what will the volume of the balloon b
Anika [276]
The answer is D. 8.1L
7 0
2 years ago
Cyclobutane, C4H8, consisting of molecules in which four carbon atoms form a ring, decomposes, when heated, to give ethylene. Th
frosja888 [35]

Answer:

The concentration of cyclobutane after 875 seconds is approximately 0.000961 M

Explanation:

The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M

The final concentration of cyclobutane, [A_t] = 0.00119 M

The time for the reaction, t = 455 seconds

Therefore, the Rate Law for the first order reaction is presented as follows;

\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }

Therefore, we get;

k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}}  {-t }

Which gives;

k = \dfrac{\text{ ln} \dfrac {0.00119}{0.00150}}  {-455} \approx 5.088 \times 10^{-4}

k ≈ 5.088 × 10⁻⁴ s⁻¹

The concentration after 875 seconds is given as follows;

[A_t] = [A₀]·e^{-k \cdot t}

Therefore;

[A_t] = 0.00150 × e^{5.088 \times 10^{-4} \times 875}  = 0.000961

The concentration of cyclobutane after 875 seconds, [A_t] ≈ 0.000961 M

6 0
3 years ago
10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of
Ilia_Sergeevich [38]

Answer:

0.355 N of HF

Explanation:

The titration reaction of HF with KOH is:

HF + KOH → H₂O + KF

<em>Where 1 mole of HF reacts per mole of KOH</em>

<em />

Moles of KOH are:

0.0296L × (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.

As volume of the titrated solution was 10.0mL, normality of HF solution is:

3.552x10⁻³ equivalents of HF / 0.010L =<em> 0.355 N of HF</em>

3 0
3 years ago
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