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Alinara [238K]
3 years ago
8

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

3 0
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.

3 0
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
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