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3 years ago

How many moles of solid naf would have to be added to 1.0 l of 2.39 m hf?

1 answer:

6 0

Molarity is one way of expressing concentration and is equal to the number of moles of the solute per liter of the solution. Therefore,

Molarity = 2.39 mol / L solution

2.39 ( 1.0) = 2.39 mol HF

Hope this answers the question.

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A) 30.22 g NaCl x 1 mol NaC l = 58.4430 Molar mass (g) NaCl=0.5171 mol NaCl

jekas [21]

The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.

So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl.

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3 years ago

One of the most worrisome waste products of a nuclear reactor is plutonium

Arte-miy333 [17]

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3 years ago

How do the first ionization energies of main group elements vary across a period and down a group?

meriva

They increase across each period, decrease down a group. As you go across a period the number of protons and increases. The positive nucleus then has a stronger attractive force on the electrons so it takes a larger amount of energy to remove an electron. As you go down a group the atoms are larger so the attractive force is weaker and it takes less energy to remove an electron.

8 0

3 years ago

(20 points) i) An absorption intensity of 1.00 (in arbitrary units) is observed for the maximum peak of the 1:2:1 triplet of the

laiz [17]

Answer:

Concentration of ethanol required = 48.476 M

Explanation:

Given that:

the absorption intensity = 1.00

Molarity of ethanol = 1M

NMR instrument used = 160 MHz

Temperature used = 300 K

The required concentration of ethanol can be determined as follows:

$= ( 1 \ M \times \dfrac{160\ MHz }{450 \ MHz}) \times \dfrac{300 \ K}{2.2\ K}$

$= ( 1 \ M \times 0.3555 ) \times136.36}$

= 48.476 M

5 0

3 years ago

Determine the empirical formula of a compound containing 40. 6 grams of carbon, 5. 1 grams of hydrogen, and 54. 2 grams of oxyge

zavuch27 [327]

The empirical formula is C₂H₃O₂

<h3>What is Empirical formula of a compound ?</h3>

The empirical formula is the simplest whole number ratio of elements present in a compound.

The total molar mass of the compound is 118.084 g/mol.

mass of Carbon present = 40.6

mass of Hydrogen present = 5.1 grams

mass of Oxygen present = 2 grams

Moles of C = 40.6/12 = 3.38

Moles of H = 5.1/1.008 = 5

Moles of Oxygen = 54.2/15.999 = 3.38

Ratio of Moles of C to Oxygen is 1 : 1

Ratio of Moles of C to H is 1/1.5

Multiplying each mole fraction by 2

The empirical formula is C₂H₃O₂

To know more about Empirical Formula

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5 0

2 years ago