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mel-nik [20]
2 years ago
12

WHAT IS THE MIXTURE OF BOILING POINT AND MELTING POINT?

Chemistry
2 answers:
mixas84 [53]2 years ago
8 0

Answer:

What is it has fixed boiling point substance or mixture? A pure substance has a sharp melting point (melts at one temperature) and a sharp boiling point (boils at one temperature). A mixture melts over a range of temperatures and boils over a range of temperatures. Homogeneous mixtures are called solutions.

Explanation:

lol they got it from go ogle

Liono4ka [1.6K]2 years ago
7 0

Answer:

A pure substance has a sharp melting point (melts at one temperature) and a sharp boiling point (boils at one temperature). A mixture melts over a range of temperatures and boils over a range of temperatures. Homogeneous mixtures are called solutions.

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After scientists use a careful process to gather evidence and establish scientific facts, which step most likely occurs next in
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Answer:

Conducting Experiments

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2 years ago
Is pre ap chemistry hard in high school?
UkoKoshka [18]

If you don't practice enough it's obviously going to be hard but if you practice enough it's going to be a piece of cake so don't think if it's going to be hard or not just think it's going to be worth the try at the very end

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3 years ago
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Find the molecular mass of.
Margaret [11]

Answer:

  1. sodium oxide Nao
  2. aluminum chloride Al2cl3
  3. sodium sulphate Naso4
  4. <em>magnes</em><em>ium</em><em> </em><em>hydroxide</em><em> </em><em>Mgoh2</em>
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3 years ago
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Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
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
Svetach [21]

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;)



4 0
2 years ago
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