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Mice21 [21]
3 years ago
5

Why do noble gases have the least amount of electronegativity?

Chemistry
1 answer:
sattari [20]3 years ago
4 0
<span>Their orbitals are completely filled</span>
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How many moles of KNO3 are in 500 mL of 2. 0 M KNO3? mol KNO3.
8090 [49]

The moles can be defined as the mass of the substance with respect to molar mass. The moles of potassium nitrate is 1 mol.

<h3>How to calculate moles of a substance?</h3>

The moles of a compound can be calculated from:

\rm Moles=\dfrac{Mass}{Molar\;mass}

The molarity can be defined as the moles of solute in a liter of solution.

The molarity can be expressed as:

\rm Molarity=\dfrac{Moles\;\times\;1000}{Volume\;(mL)}

The molarity of potassium nitrate solution is 2 M, and the volume is 500 mL.

The moles of potassium nitrate is given as:

\rm 2\;M=\dfrac{Moles\;\times\;1000}{500\;mL}\\ Moles=\dfrac{2\;\times\;500}{1000}\;mol\\ Moles=1\;mol

The moles of potassium nitrate in 2 M, 500 mL solution are 1 mol.

Learn more about moles, here:

brainly.com/question/15209553

4 0
1 year ago
If 2.0 mL of 6.0M HCl is used to make a 500.0-mL aqueous solution, what is the molarity?
RideAnS [48]

Answer:

0.024M

Explanation:

Data obtained from the question include:

C1 = 6M

V1 = 2mL

C2 =?

V2 = 500mL

The molarity of the diluted solution can be obtained as follows:

C1V1 = C2V2

6 x 2 = C2 x 500

Divide both side by 500

C2 = (6 x 2) /500

C2 = 0.024M

The molarity of the diluted solution is 0.024M

5 0
3 years ago
Read 2 more answers
Why is a liquid able to flow
luda_lava [24]
Since the particles can move, the liquid<span> can flow</span> <span>and take the shape of its container. Some insects, such as pond-skaters, are able</span><span> to walk on water without sinking. This is because the forces of attraction between the water particles pull the particles at the surface together.</span>
4 0
3 years ago
Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engin
OlgaM077 [116]

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}

The temperatures have to be expressed in Rankine (or Kelvin) degrees

1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F

(b) The Carnot efficiency of the cycle is

\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502

If the efficiency of the plant is 60% of the Carnot efficiency, we have

\eta=0.6*\eta_{c}=0.6*0.502=0.302

The heat used in the plant can be calculated as

Q_i=W/\eta=750MW/0.302=2483MW

And the heat removed to the heat sink is

Q_o=Qi-W=2483-750=1733MW

If the flow of the river is 165 m3/s, the heat per volume in the sink is

\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C

8 0
3 years ago
Which of the following have at least one cell?
Zarrin [17]

Answer: all of the answers except for rocks. Glad to help :D need help with anything else?

7 0
2 years ago
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