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grin007 [14]
2 years ago
11

What is the molarity of 1. 5 liters of an aqueous solution that contains 52 grams of lithium fluoride.

Chemistry
1 answer:
Cerrena [4.2K]2 years ago
8 0

Answer:

1.3 M

Explanation:

The formula mass of lithium fluoride is 7+19=26 g/mol. This means that in 52 grams of LiF, there is 2 moles.

Molarity = (moles of solute)/(liters of solution), so the answer is 2/1.5 = <u>1</u><u>.</u><u>3</u><u> </u><u>M</u>

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Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas:NaH (s) H2O (l) → NaOH (aq) H2 (g)A
Anuta_ua [19.1K]

NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)

You want to calculate the mass of NaH, I assume.  Otherwise, the question isn't clear.  It simply says calculate the mass(??)

 

So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr.  But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas.  So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.

PV = nRT

n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)

n = 0.0385 moles H2

 

moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required

mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH

Brainliest Please :)

7 0
3 years ago
Read 2 more answers
If element x forms the oxides xo and x2o3 the oxidation numbers of element x are
Deffense [45]

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

XO

And:

X_2O_3

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

X^{+2}O^{-2}

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

X_2^{+3}O_3^{-2}

Thus, answer is b)+2 and +3

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3 0
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wlad13 [49]

Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).

Explanation:

This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7

Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4

3 0
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Hi, can someone help me with chemistry
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Question 19 of 26
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Explanation:

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