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puteri [66]
3 years ago
11

What is an atom??????????

Chemistry
2 answers:
FromTheMoon [43]3 years ago
7 0
<span>the basic unit of a chemical element.</span>
Viktor [21]3 years ago
7 0
<span>The basic unit of a chemical element </span>and a extremely small amount of a thing or quality.

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5.00 g of an unknown compound was dissolved in 250.0 mL of water. The unknown compound was found to be a non-electrolyte and the
Zielflug [23.3K]

Answer:

This unknown compound has a molar mass of 33.74g/mole

Explanation:

<u>Step 1:</u> Given data

The formula for the osmotic pressure is:

π = MRT = 14.5 atm

⇒ with M = Concentration of the solution

⇒ with R = gas constant = 0.08206 L atm / mol K

⇒ with T = absolute temperature (in Kelvin) = 298K

<u>Step 2:</u> Calculate concentration

M = π/ RT

M = 14.5 / (0.08206 * 298)

M = 0.593 M this is 0.593 moles per L

<u>Step 3:</u> Calculate number of moles

Since the volume is only 0.250 L, the amount of moles is:

0.593 * 0.250 L = 0.1482 moles

<u>Step 4:</u> Calculate molar mass

The molar mass is : mass / moles

Molar mass of this unknown compund = 5g / 0.1482 moles = 33.74 g/mole

This unknown compound has a molar mass of 33.74g/ mole

4 0
3 years ago
Identify which of the concentration expressions can also be used to describe a solution with a concentration of 1 mg/mL of solut
Alexxandr [17]

Answer: Parts per million (ppm)

Explanation:

Consider the units milligram per milliliter. This gives us one part of the solute per one million parts of solvent. That is 10^ -3/10^-3= 10^-6. This unit is commonly used in analytical chemistry to show very small concentration of analyte. A similar unit is parts per billion(ppb)

3 0
4 years ago
Specify the l and ml values for n = 4.
Bumek [7]

Answer : The specify the l and ml values for n = 4 are:

At l = 0,  m_l=0

At l = 1,  m_l=+1,0,-1

At l = 2,  m_l=+2,+1,0,-1,-2

At l = 3,  m_l=+3,+2,+1,0,-1,-2,-3

Explanation:

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from (-l\text{ to }+l). When l = 2, the value of m_l will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as m_sThe value of this is +\frac{1}{2} for upward spin and -\frac{1}{2} for downward spin.

As we are given, n = 4 then the value of l and ml are,

l = 0, 1, 2, 3

At l = 0,  m_l=0

At l = 1,  m_l=+1,0,-1

At l = 2,  m_l=+2,+1,0,-1,-2

At l = 3,  m_l=+3,+2,+1,0,-1,-2,-3

6 0
3 years ago
If 65.5 ml of HCl stock solution is used to make 450 ml of a 0.675 M HCl dilution, what is the molarity of the stock solution ?
Talja [164]

<u>Answer:</u> The molarity of the stock solution is 4.64 M

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock solution

M_2\text{ and }V_2 are the molarity and volume of diluted solution

We are given:

M_1=?M\\V_1=65.5mL\\M_2=0.675M\\V_2=450mL

Putting values in above equation, we get:  

M_1\times 65.5=0.675\times 450\\\\M_1=4.64M

Hence, the molarity of the stock solution is 4.64 M

8 0
3 years ago
Determine the partial negative charge on the bromine atom in a c−br bond. the bond length is 1.93 å and the bond dipole moment i
vaieri [72.5K]

Answer:

The value is x  =  0.151  \ e

Explanation:

From the question we are told that

The bond length is l  =  1.93\  \r  a =  1.93 *1 *10^{-10}  =1.93 *10^{-10}\  m

The bond dipole moment is \mu  = 1.40 d  = 1.40 *  3.33564 *10^{-30}  =  4.6699 *10^{-30} \  C \cdot m

Generally the dipole moment is mathematically represented as

\mu  =  Q *  l

Here Q is the partial negative charge on the bromine atom

So

Q =  \frac{\mu}{ l}

=> Q =  \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }

=> Q = 2.42 *10^{-20} C

Generally

1 electronic charge(e) is equivalent to 1.60*10^{-19} C

So x electronic charge(e) is equivalent to Q = 2.42 *10^{-20} C

=> x  =  \frac{2.42 *10^{-20}}{1.60*10^{-19} }

=>     x  =  0.151  \ e

3 0
3 years ago
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