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Lunna [17]
3 years ago
9

An unknown compound in the lab is either LiOH, CaF2, or LiF. If the sample has a mass of 260 grams and contains 3.33 moles, what

is the identity of the compound?​
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

CaF2

Explanation:

1 mole of CaF2 = 78.0748064 grams

78 times 3.33= 259.74

the others multiplied were nowhere near 260 grams either

You might be interested in
PLEASE HELP!<br><br> See picture
Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

5 0
3 years ago
What volume would 0.853 moles of Nitrogen gas occupy at STP?
beks73 [17]

Answer:

19.12 L

Explanation:

At STP(i.e. Standard temperature and pressure).

The volume occupied by one mole of gas = 22.4 L

The pressure = 1 atm

The temperature = 273 K

Thus, since 1 mole of gas = 22.4 L;

Then 0.853 moles of N2 gas will occupy:

= (0.853 moles of N2 gas × 22.4 L)/ 1 mole of N2 gas

= 19.12 L

6 0
3 years ago
Maria took a walk to the store, did some shopping, and then came home. During Maria's trip, when was her displacement equal to z
velikii [3]
I would say it would be a. It makes the most sense
6 0
3 years ago
If D+2 would react with E-", what do you predict to be the formula?
GuDViN [60]

Answer:DE2

Explanation:

6 0
3 years ago
N204(0) + 2NO2(g)
user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state  

Q is the product of the ion ions from the reacting substance  

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}

Setup 1 :

\tt Q=\dfrac{0.0064^2}{0.098}=0.000418=4.18\times 10^{-4}

Q<K⇒The reaction moved to the right (products)

Setup 2 :

\tt Q=\dfrac{0.0304^2}{0.15}=0.00616=6.16\times 10^{-3}

Q=K⇒the system at equilibrium

Setup 3 :

\tt Q=\dfrac{0.230^2}{0.420}=0.126

Q>K⇒The reaction moved to the left (reactants)

8 0
2 years ago
Read 2 more answers
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