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Genrish500 [490]
2 years ago
7

One of the goal's of an atom is:

Chemistry
1 answer:
iren [92.7K]2 years ago
5 0

Answer:

to have a "full" last orbital shell

Explanation:

They can not live withought ther outer most shell

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How many neutrons are in a single atom of indium-115?
anzhelika [568]
Indium has 49 protons 

mass number (# of neutrons and # of protons combined) is 115

115 - 49 = 66

66 neutrons
8 0
3 years ago
When light mass nuclei combine to form a heavier more stable nucleus this is called​
n200080 [17]

Answer:

Nuclear fusion

Explanation:

Science

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3 years ago
Use the portion of the periodic table shown below to answer the questions
leva [86]

Answer:

Part 1:sodium

rubidium

Part 2: protons neutrons and electrons are all 12

The number of protons is equal to the no. of neutrons from the electronic arrangement of magnesium and the no. of electrons is got from the atomic no. of magnesium

8 0
2 years ago
3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con
KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}

c) Half life for first-order kinetics is computed by:

t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}}  =53.3min

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M

Best regards.

6 0
3 years ago
Can someone please help me with 363?
Hitman42 [59]

htdtshhthtdhthshAnswer:

yyrgtyht5gggggggggggggggggdggdd

Explanation:

6 0
2 years ago
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