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brilliants [131]
2 years ago
7

Fire a single electron towards the hydrogen atom. Describe what happens in a step by step fashion. [N.B. - It may be helpful to

utilize the Run in Slow Motion option for this part.]
Chemistry
1 answer:
scZoUnD [109]2 years ago
7 0

Answer:

The electrons will be added by the hydrogen.

Explanation:

If we fire a single electron towards the hydrogen atom, the hydrogen atoms added the electron to its shell by applying force of attraction and becomes stable as well as non reactive in nature because the hydrogen attains the electronic configuration of helium which is a noble gas and have completed its outermost shell. The proton that is present in the nucleus attracts this electron and compel it to add in the electron.

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1. Determine which object will SINK or FLOAT in water.
zubka84 [21]

Answer: Block # 1 , Block # 4 and

Block # 5 will sink.

Block # 2 and 3 will float.

Explanation: The density of water is equal to 1 g/mL. Any density that is less than the density of water will float. Objects with higher density compared to water will eventually sink.

8 0
3 years ago
How many moles are in 297g of nh3?<br><br> Please show work, will give brainliest.
rusak2 [61]

Answer:

17.5moles

Explanation:

The number of moles in a substance can be calculated by using the formula;

Number of moles (n) = mass (m) ÷ molar mass (MM)

According to this question, mass of ammonia (NH3) = 297g

Molar Mass of NH3 = 14 + 1(3)

= 17g/mol

n = 297/17

n = 17.47

Number of moles of NH3 = 17.5moles

3 0
3 years ago
How many milliseconds are there in 3.5 seconds
azamat
There are 3,500 milliseconds in 3.5 seconds. Hope this helps!
5 0
3 years ago
Read 2 more answers
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

6 0
3 years ago
When a gas has a pressure of 1.3 atm at a volume of 34 ml what would the volume be if the pressure was increased to 1.7 atm
aleksandr82 [10.1K]

Answer:

unknown pressure has its volume increased to 34 liters and its temperature ... atm, what will be the pressure of the gas if I raise the temperature to 94.

5 0
3 years ago
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