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shutvik [7]
3 years ago
7

Can someone please help me with these questions. image attached

Chemistry
1 answer:
Radda [10]3 years ago
7 0
Question 9. The first one is the smallest. Anything with a negative exponent is going to be less than 1, the .00000241. The exponent tells you the number of zeroes to the right of the decimal point. Farther to right gets smaller and smaller.

Question 10. The last one is true. If the last digit is smaller than 5, drop the digit, and do not change. (If it is a 5 or larger, the digit before it would round up)
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A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what
bulgar [2K]

Answer:

The mass of KF in the mixture is 2.77 gms.

Explanation:

Given;

Total weight of mixture (LiF+KF)=5.32gms

Let, mass of KF in the mixture = x gms

⟹ mass of LiF in mixture =(5.32-x)gms.

We know that :

Atomic weight of F=19gms.

Atomic weight of Li =7gms.

Atomic weight of K = 39 gms.

moles=mass/(molecular weight)

Thus, moles of KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

Thus,

moles of F in KF=moles of KF=x/58 ---(1)

moles of F in LiF =moles of LiF= (5.32-x)/26---(2)

From (1) & (2),

Total moles of Fluorine

=(x/58)+((5.32-x)/26)

Hence,

total weight of Fluorine in sample = moles*Atomic weight

=((x/58)+((5.32-x)/26))*19gms.

=3.12 gms.---(given)

Now, solving the equation for x,

26x +(5.32*58)-58x

=3.12*58*26/19

22x=308.56-247.62

x=60.94/22

=2.77 gms. (Answer)

Thus, the mass of KF in the mixture is 2.77 gms.

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idk

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