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bekas [8.4K]
2 years ago
5

How many oxygen atoms are present in a 14.0 g sample of Cu(NO3)2

Chemistry
1 answer:
Vikentia [17]2 years ago
4 0
<span>3 x 2 =6 oxygen atoms and multiply 6 by 14.0 and you have an answer. </span>
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The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close
Natasha_Volkova [10]

Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log(A^-/HA) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

6 0
2 years ago
Which runner has greater kinetic energy: a 46-kilogram runner moving at a speed of 8 meters per second or a 92-kilogram runner m
Lena [83]
Runner a has a greater kinetic energy
8 0
2 years ago
Freon-11, CCl3F has been commonly used in air conditioners. It has a molar mass of 137.35 g/mol and its enthalpy of vaporization
PilotLPTM [1.2K]

Answer:

180.56 Kilo joules of energy is removed in the form of heat when 1.00 kg of freon-11 is evaporated.

Explanation:

Molar mass of freon-11 = 137.35 g/mol

Enthalpy of vaporization of freon-11= \Delta H_{vap}=24.8 kJ/mol

Mass of freon-11 evaporated = 1.00 kg = 1000 g

Moles of freon-11 evaporated = \frac{1000 g}{137.35 g/mol}=7.2807 mol

Energy in the form of heat removed when 1.00 kg of freon-11 gets evaporated:

7.28067 mol\times \Delta H_{vap}=7.2807 mol\times 24.8 kJ/mol

=180.56 kJ

0 0
2 years ago
How many moles of Helium gas are in a balloon with a volume of 6.452 L at 99.7 kPa
Taya2010 [7]

Answer:

n = 0.26 mol.

Explanation:

Given,

Pressure, P = 99.7 kPa = 1 atm

where 101.325 kPa  = 1 atm

            P = 0.984 atm

Temperature, T = 297 K

Volume = 6.452 L

Now, using ideal gas equation

PV = n RT

0.984 x 6.452 = n x 0.08206 x 297

n = 0.26 mol.

8 0
2 years ago
How many atoms are in 80.45 g of magnesium?
KatRina [158]

Hello!

To find the amount of atoms that are in 80.45 grams of magnesium, we will need to know Avogadro's number and the mass of one mole of magnesium.

Avogadro's number is 6.02 x 10^23 atoms, and one mole of magnesium is equal to 24.31 grams.

1. Divide by one mole of magnesium

80.45 / 24.31 = 3.309 moles (rounded to the number of sigfigs)

2. Multiply moles by Avogadro's number

3.309 x (6.02 x 10^23) = 1.99 x 10^24 (rounded to the number of sigfigs)

Therefore, there are 1.99 x 10^24 atoms in 80.45 grams of magnesium.

8 0
2 years ago
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