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Vlada [557]
3 years ago
10

A sample of a compound containing boron (B) and hydrogen (H) contains 5.443 g of B and 1.522 g of H. The molar mass of the compo

und is about 30 g. What is its molecular formula?
Chemistry
1 answer:
Svetach [21]3 years ago
4 0

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

first you divide the mass of BORON by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass.

So, 5.443/11 is about 0.5. Then, the RFM of H is 1, so the number of mole is 1.522/1=1.522.

Next, you get the number of moles in order is; 0.5 and 1.522. Now we need to look at the ratio between these numbers. 0.5 is smaller so we use it as the ratio of 1.  next use 1.522/0.5 is 3.044 which has a greatest common factor of 3. so the empirical formula is BH3.

Now we are going to solve the molecular formula.

the molar mass ofthis compound is 30g, so we're going to find the RFM of the empirical formulsof BH3 first.

11+3=14.

now we see how many times 14 goes into 30. 30/14=2.14 which is about 2.

So now we need to times the subscript of the empirical formula by two.

thus, the molecular formula is B2H6.


To solve this kind of  questions, there are many steps:Know what you are calculating about, it's about the molecular formula, so you need to find out the number of moles of each elements. then use the molar mass of the whole compound to calculate the molecular formula.

1) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4) then use the molarmass of the whole compound to find the molecular formula. molar mass of the whole compound/RFM(molar mass) of the empirical formula of elements= the number you need to multiply by the subscript of the empirical formula to get the molecular formula.

please tell me if i got anything wrong;)



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Consider the following operations on the number 7.02 times 10^-2. Without using a calculator, decide which would give a signific
mariarad [96]

Answer: (a). Significantly Larger value

(b). Significantly Smaller value

(c). Significantly Larger value

(d). Significantly Smaller value

Explanation:

This is quite a dicey  question to analyze, let us not get too carried away with the simplicity of the mathematical signs involved.

Given that 7.02 × 10⁻² is the number to be compared with.

(a).  7.02 × 10⁻² + 6.10 × 10⁵

from this, we can see a very large number being added to the base number involved, although the base number is carrying a negative power, it does not affect the other greatly.

ANS: the sum would give a significantly larger value

(b). 7.02 × 10⁻² - 6.10 × 10⁵

Here, we can see a very large number subtracted from the base number having a negative power.

the subtraction of this numbers gives a Significantly smaller value than the base number i.e. 7.02 × 10⁻²

ANS: Significantly smaller value

(c). 7.02 × 10⁻² × 6.10 × 10⁵

we would solve this from our basic knowledge of indices

where ⇒ (10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ)

we have,

7.02 × 10⁻² × 6.10 × 10⁵ = (7.02 × 6.10) × 10⁻²⁺⁵ = (7.02 × 6.10) × 10³

this final value (multiplication value) gives a Significantly larger value.

ANS: Significantly larger value

(d). 7.02 × 10⁻² ÷ 6.10 × 10⁵

Also, we apply the indices rule for Division

where ⇒ (10ᵃ ÷ 10ᵇ = 10ᵃ⁻ᵇ)

i.e. 7.02 × 10⁻² ÷ 6.10 × 10⁵ = (7.02 × 6.10) × 10⁻²⁻⁺⁵ = (7.02 × 6.10) × 10⁻⁷

this final value gives a Significantly Smaller value

ANS: Significantly Smaller value

cheers i hope this helps.

7 0
3 years ago
A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25
mylen [45]
<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

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We are required to calculate the volume of diluted solution;

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M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

     = (1.00 M × 50 mL) ÷ 0.25 M

     = 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

5 0
3 years ago
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