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Marina CMI [18]
3 years ago
9

When an electron is removed the atoms charge changes to a what?

Chemistry
1 answer:
inna [77]3 years ago
6 0

Answer:

It make the atom octet from i.e more stable form.

Explanation:

A neutral sodium atom is likely to achieve an octet in its outermost shell by losing its one valence electron. Na -----> Na+ e-

You might be interested in
Consider the generic reaction: 2 A(g) + B(g) → 2 C(g). If a flask initially contains 1.0 atm of A and 1.0 atm of B, what is the
irina1246 [14]

Answer:

b. 1.5 atm.

Explanation:

Hello!

In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:

P=P_A+P_B+P_C

Now, if we introduce the stoichiometry, and the change in the pressure x we can write:

P=1.0-2x+1.0-x+2x

Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:

x=\frac{1.0atm}{2}=0.5atm

Thus, the final pressure is:

P=1.0-2(0.5)+1.0-(0.5)+2(0.5)\\\\P=1.5atm

Therefore the answer is b. 1.5 atm.

Best regards!

3 0
3 years ago
A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud
Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

                             = 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence,   Mass = gcm-3 * cm3

              Mass = g

The mass of pentane the student should weigh out is 28.17g

Learn more about Density on

brainly.com/question/1354972

#SPJ1

5 0
2 years ago
Help me please byubflyfheqwucheuqcbidlvbyeqvygieuqvuleqvhyiroev
Reptile [31]

Answer:

The answer you have selected is correct

Explanation:

4 0
3 years ago
A 312 g sample of a metal is heated to 277.845 °C and plunged into 200 g of water at a temperature of 11.945 °C. The final tempe
tigry1 [53]

Answer:

The specific heat capacity of the metal is 1.307 J/g °C

Explanation:

<u>Step 1:</u> Data given

mass of the metal = 312 grams

initial temperature of the metal ( before plunged in the water) = 277.845 °C

initial temperature water = 11.945 °C

Final temperature of water (and aslo the metal) = 99.062 °C

Specific heat capacity of wayer = 4.184 J/g °C

<u>Step 2:</u>  Calculate the specifi heat capacity of the metal

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(metal) * C(metal) * ΔT(metal) = - m(water) * C(water) * ΔT(water)

with mass of metal = 312 grams

with C(metal) = TO BE DETERMINED

with ΔT (metal) = 99.062 - 277.845 = -178.783 °C

with mass of water = 200 grams

with C(water) = 4.184 J/°C * g

with ΔT (water) = 99.062 - 11.945 = 87.117 °C

312 * C(metal) * -178.783  = - 200* 4.184 * 87.117

C(metal) = -72899.51 / 55780.296 = 1.307

The specific heat capacity of the metal is 1.307 J/g °C

6 0
3 years ago
A sample of oxygen gas at a pressure of 690 mmHg and a temperature of 32°C occupies a volume of 14.2 L if the gas is cooled at c
Diano4ka-milaya [45]

Answer:

12.66 L

Explanation:

- Used combined gas law formula

- Change C to K

- Rearrange formula to fit your values

- If you would like further explanation or want to learn how to do these please let me know. Hope this helped!

5 0
4 years ago
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