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Tresset [83]
2 years ago
11

Help please! I don’t understand this

Chemistry
1 answer:
AlexFokin [52]2 years ago
5 0

I don’t really know how to explain how to write electron configurations in a short paragraph so maybe look up a video? It’s really easy to understand once you get the hang of it but i’ll attach a picture to help you write electron configurations for any element. Try to think of it as reading the periodic table from left to write until you end up on the element.

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The molarity of an aqueous sodium phosphate solution is 0.650 M. What is the molality of sodium ions present in this solution? T
netineya [11]

Answer:

molality of sodium ions is 1.473 m

Explanation:

Molarity is moles of solute per litre of solution

Molality is moles of solute per kg of solvent.

The volume of solution = 1 L

The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams

The mass of solute = moles X molar mass of sodium phosphate = 0.65X164

mass of solute = 106.6 grams

the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg

the molality = \frac{moles of solute}{mass of solvent in kg}=\frac{0.65}{1.323}= 0.491m

Thus molality of sodium phosphate is 0.491 m

Each sodium phosphate of molecule will give three sodium ions.

Thus molality of sodium ions = 3 X 0.491 = 1.473 m

7 0
3 years ago
What atom has the largest atomic radius
lesya692 [45]

Answer:

Caesium

down the group atomic radius increases and along period wise atomic radius decreases

5 0
3 years ago
Read 2 more answers
Give the name of the following molecule. A compound with a total of six carbons with a double bond and the rest single bonds. Th
AlekseyPX
Answer:
            The name of given molecule is 3-Methylpent-2-ene.

Explanation:
                   First of all a carbon chain of five carbons was drawn. Then a double bond was made between carbon 3 and 4 (starting from left). A methyl group was drawn at middle carbon which is at position 3.

                    Molecule sketched was named as,

1) A longest chain containing double bond was selected and numbering was started from the end closest to double bond. Hence, 

                                  2-Pentene or Pent-2-ene

2) The position of substituent was specified before the parent name, Hence,

                       3-Methyl-2-Pentene or 3-Methylpent-2-ene

7 0
2 years ago
0.20 L of HNO3 is titrated to equivalence using 0.12 L of 0.2 M NaOH. What is the concentration of the HNO3?
AleksandrR [38]

M_{A}V_{A}=M_{B}V_{B}\\(0.20)(M_{A})=(0.12)(0.2)\\M_{A}=\frac{(0.12)(0.2)}{(0.20)}=\boxed{0.12 \text{ M}}

4 0
1 year ago
A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an
inysia [295]

<u>Answer:</u> The molar mass of the unknown compound is 223.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 8.44 torr

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 62.3637\text{ L torr }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 4.54\times 10^{-4}M

Given mass of unknown compound = 15.2 mg = 0.0152 g   (Conversion factor:  1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol

Hence, the molar mass of the unknown compound is 223.2 g/mol

5 0
3 years ago
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