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SOVA2 [1]
3 years ago
8

How does an atom change if all of its electrons are removed?

Chemistry
2 answers:
amm18123 years ago
6 0
Adding or removing<span> them from the nucleus </span>does<span> not</span>change<span> the electrical charge of the nucleus.</span>
Dafna1 [17]3 years ago
3 0
If all the electrons are removed, the atom becomes a cation because it is a negative charged atom.
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Oceanic crust is
Ratling [72]

Answer:

It is thinner than the continental crust.

Explanation:

4 0
3 years ago
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How much energy in joules does it take to raise the temperature of 1.5kg of aluminum from 20c to 40c
Leviafan [203]

Answer:

Q = m * c * ΔΤ

ΔΤ = 40-20 = 20

Q = 1.5kg * 0.9 * 20

Q = 27j

4 0
3 years ago
Describe the effect of subjecting hydrogen to pressure​
ExtremeBDS [4]

Answer:

When hydrogen is subjected to large enough pressure, it solidifies according to theory.

Explanation:

According to theory, when  hydrogen molecules are subjected to enormous degree of pressure the molecules will solidify.

What happens here is that the hydrogen–hydrogen bonds in the hydrogen molecule will break apart and the molecules collapses into hydrogen atoms.

Hence, when hydrogen is subjected to large enough pressure, it solidifies according to theory.

3 0
3 years ago
Calculate the temperature required to make a 75.8 ml ballon initially at 57.9c change its value to 50.8ml
nika2105 [10]

Answer:

T₂ = 221.8 K

Explanation:

Given data:

Initial temperature = 57.9°C ( 57.9+273 =330.9 k)

Initial volume = 75.8 mL

Final volume = 50.8 mL

Final temperature = ?

Solution:

According to Charles's law,

V₁ /T₁= V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 50.8 mL ×330.9 k / 75.8 mL

T₂ = 16809.72 mL.K/ 75.8 mL

T₂ = 221.8 K

6 0
4 years ago
All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou
____ [38]

Answer:

m_{Mg}=30.8mgMg

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg

Best regards.

3 0
3 years ago
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