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tigry1 [53]
3 years ago
7

What is the very small center core of an atom is called?

Chemistry
2 answers:
Tanzania [10]3 years ago
5 0
<span>The very small center core of an atom is called</span> the nuclues, where the protons and neutrons are located. 
madam [21]3 years ago
4 0
The small part in the center of the atom is called a nucleus
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How many f orbitals have the value n = 3?
dalvyx [7]
The Answer You Are Looking For Is.......


b) 3



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4 0
3 years ago
What determines crystal size in minerals formed by lava or magma?
Burka [1]

Answer:

When magma cools, crystals form because the solution is super-saturated with respect to some minerals. If the magma cools quickly, the crystals do not have much time to form, so they are very small. If the magma cools slowly, then the crystals have enough time to grow and become large.

Explanation:

5 0
2 years ago
A balloon of helium gas in Alabama has a volume of 22.4 L at 26°C and a 1.03 ATM how big with a balloon get if is transported to
Dimas [21]

Answer:

The final volume of the balloon is = 28.11 L

Explanation:

Initial pressure P_{1} = 1.03 atm = 104.325 K pa

Initial temperature T_{1} = 26 °c = 299 K

Initial volume V_{1} = 22.4 L

Final temperature T_{2} = 22 °c = 295 K  

Final pressure P_{2} = 0.81 atm = 82 K pa

We know that

\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }

Put all the values in above formula we get

\frac{(104.325)(22.4)}{299} = \frac{(82)(V_{2} )}{295}

V_{2} = 28.11 L

This is the final volume of the balloon.

8 0
3 years ago
Xian drew a diagram to compare deletion mutations and substitution mutations. Which label belongs in the area marked "X"? may ch
tankabanditka [31]

Answer: (B)

Explanation: Decreases the number of bases in the sequence.

3 0
3 years ago
Read 2 more answers
Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enth
stealth61 [152]

Answer : The value of \Delta H_{vap} is 28.97 kJ/mol

Explanation :

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature -21.0^oC = 462.7 mmHg

P_2 = vapor pressure at temperature -44.0^oC = 140.5 mmHg

\Delta H_{vap} = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = -21.0^oC=[-21.0+273]K=252K

T_2 = final temperature = 45^oC=[-41.0+273]K=232K

Putting values in above equation, we get:

\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol

Therefore, the value of \Delta H_{vap} is 28.97 kJ/mol

4 0
3 years ago
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