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Agata [3.3K]
3 years ago
9

in this cut-away image of an atom, where is the red arrow most likely pointed at the greatest probability of locating an electro

n

Chemistry
2 answers:
iren [92.7K]3 years ago
5 0
It’s C :) hopefully its right ^_^
emmasim [6.3K]3 years ago
3 0

Answer:

c

Explanation:

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9A. A sample of hydrogen at 1.56 atm had it's pressure decreased to 0.73
densk [106]

Answer:

351.43mL

Explanation:

To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:

P1V1 = P2V2

Where; P1 = initial pressure

V1 = initial volume

P2 = final pressure

V2 = final volume

According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL

Hence;

P1V1 = P2V2

1.56 × V1 = 0.73 × 751

1.56 V1 = 548.23

V1 = 548.23/1.56

V1 = 351.43mL

Therefore, the original volume of hydrogen gas is 351.43 mL.

4 0
4 years ago
The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of
V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, 1401.5kJ(\frac{1kcal}{4.184kJ})

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.


8 0
3 years ago
Please help with the question attached<br>I need it in an hour​
lawyer [7]

Answer:

iiiiiiddddiiilllooeiuehe

4 0
2 years ago
A student decides to conduct an investigation to determine the boiling points of water and ethanol. The student will heat sample
lilavasa [31]
Deciding on the appropriate equipment
8 0
3 years ago
A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate
Afina-wow [57]

Answer:

136.63 °C

Explanation:

ΔTb=Tb solution - Tb pure

Where; Tb pure = 133.60°C

molar mass of solute = 121.14 g/mol

number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles

molality = 0.431 moles/350 * 10^-3 = 1.23 molal

Then;

ΔTb = Kb * m * i

Kb = 2.46°C kg mol^-1

m = 1.23 molal

i = 1

ΔTb = 2.46 * 1.23 * 1

ΔTb = 3.03 °C

Hence;

Tb solution = ΔTb + Tb pure

Tb solution = 3.03 °C + 133.60°C

Tb solution = 136.63 °C

6 0
3 years ago
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