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3 years ago

What is the density of and object that is 100 grams and has a volume of 5 cubic centimeters?

Chemistry

2 answers:

kenny6666 [7]3 years ago

7 0

Answer:

J. 20 g/cm^3

Explanation:

The first answer should be g/cm^3, not g/cm

lesya [120]3 years ago

5 0

Explanation:

Density of object = gram/volume

= 100 g/ 5 cm³

= 20 g/cm³

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If the volume occupied by the gas molecules shown below were doubled, what would happen to the pressure they exert? (Assume cons

AnnZ [28]

there's no picture here but I guess the answer would be:

considering the constant temperature, if you double the volume, the pressure would be halved.

like: volume is 2, pressure is 4

if 2×2, then:4÷2

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3 years ago

Part E

AlekseyPX

Answer: similar, energy

Explanation: (edmentum answer) Several of the elements in period 5 have electron configurations with only one electron in the 5s sublevel. That suggests that the next sublevel that electrons typically fill, 4d, is not much higher in energy.

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3 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

Write with example of any three changes in the state of matter in daily life

Stella [2.4K]

Answer:

just wait 3 minutes ill tell you the answer cause its in my book

Explanation:

7 0

3 years ago

How many milliliters of 0.0510 m edta are required to react with 50.0 ml of 0.0200 m cu2 ?

krok68 [10]

Cu ions plus EDTA2- ->cu(EDTA)2- plus 2H-
number of moles of CU ions used which is equal to molarity multiplied by volume in litres
that is 50xo.o2 divided by 1000 that is 0.001moles
Since ratio is 1:1 the moles of EDTA is also0.001moles
volume of EDTA is 0.001 divided by 0.0510m which is 0.0196 litres in ml is 0.0196x1000 which is 19.61ml

4 0

3 years ago