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2 years ago

Which is the symbol for the element that contains six electrons in each of its neutral atoms?

Chemistry

1 answer:

Harlamova29_29 [7]2 years ago

8 0

the symbol for the elements that contains six rlectrons in each of its neutrual atoms are number of 6

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The specific heat for liquid ethanol is 2.46 J/(g•°C). When 210 g of ethanol is cooled from

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Answer:

a) heat it from 23.0 to 78.3

q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =

b) boil it at 78.3

(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =

c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ

Explanation:

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3 years ago

Convert 0.0598 to scientific notation

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A sample of hydrogen is collected in a flask

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3 years ago

An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH . The pKa

fgiga [73]

Answer: The pH of the solution is 12.61

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(1)

For hydrazoic acid:

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

$1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol$

For NaOH:

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

$0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol$

The chemical reaction for hydrazoic acid and NaOH follows the equation:

$HN_3+NaOH\rightarrow NaN_3+H_2O$

Initial: 0.291 0.342

Final: 0 0.051 0.291 0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L (Conversion factor: 1 L = 1000 mL)

For NaOH left:

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M$

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.0408M$

Putting values in above equation, we get:

$pOH=-\log(0.0408)\\\\pOH=1.39$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\\\pH=14-1.39=12.61$

Hence, the pH of the solution is 12.61

8 0

2 years ago