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Nutka1998 [239]
3 years ago
14

1. Describe how you can use the graduated cylinder to determine the volume of a small, irregularly-shaped metal. 2. What safety

precaution should you take when adding the irregularly-shaped metal into the glass graduated cylinder
Chemistry
1 answer:
alisha [4.7K]3 years ago
7 0

Explanation:

1.

i. Measure the initial volume of water in a graduated cylinder.

ii. Now submerge the irregularly shaped metal into water in a graduated cylinder.

iii. Measure the final volume of water.

iv. The difference between the final volume and the initial volume is the volume of the irregularly shaped metal.

2.

When adding the irregularly shaped metal into the glass graduated cylinder. Make sure that the irregularly shaped metal is completely submerged into the water.  

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Explanation:

A. 55

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Please HELP, ASAP! Homework due tomorrow and I need some help.
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How to find the number of neutrons in an isotope?
Alexeev081 [22]
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3 years ago
How much water will be required to dissolve 0.3g of benzoic acid at 95oC given that the solubility of BA at 95oC is 68.0g/L?
Ainat [17]
Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)

Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g

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6 0
2 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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