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3 years ago

Question 1 1 pts How many mols of bromine are present in 35.7g of Tin(IV) bromate?

Chemistry

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

$n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol$

So, there are 0.0814 moles of bromine in 35.7g of Tin(IV) bromate.

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Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to

hram777 [196]

Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

$MCl_n=MCl_2$

In that compartment $[M^{n+}]=[m^{2+}]=8.279 \ M$

The volume of the $MCl_n$ taken in that compartment = 6.380 mL

So, the number of moles of $M^{2+} = 8.279 \times 6.380$

= 52.82 m mol

= 0.05280 mol

$MCl_n \rightarrow M^{n+}+nCl^-$

But n = 2

Therefore, moles of $Cl^-$ = 2 x moles of $M^{n+}$

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3 years ago

Draw the dot structure of CX4

hichkok12 [17]

Refer to the attachment

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3 years ago

A compound is 40.0% c, 6.70% h, and 53.3% o by mass. assume that we have a 100.-g sample of this compound. the molecular formula

atroni [7]

When you have 100 g of compound, then based on the percentages given, there are 40.0 g C, 6.70 g H, and 53.3 g O. Convert those to moles:

C: 40.0 g / 12.0 = 3.33 moles of C
H: 6.70 g / 1.01 = 6.63 moles of H 
O: 53.3 / 16.0 = 3.33 moles of O

Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1
So, The empirical formula is CH2O.
Now, That formula has a molar mass of [12.0 + 2(1.0) + 16.0] = 30.0

And we are given it's molar mass is = 240

So, no. of units of CH2O = 240 / 30 = 8

8 x CH2O = C8H16O8, and that is the molecular formula.

$C_8H_{16}O_8$

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Answer:

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