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Mama L [17]
4 years ago
13

PLZ HELP

Chemistry
2 answers:
nika2105 [10]4 years ago
8 0
Mjdksksksnsjdkjdjsksksjdndnd
denis23 [38]4 years ago
3 0

I do not believe you can tell the displacement by the information given. so i would say none of the above.

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What are twin primes between 1 to 100<br>​
Furkat [3]
The twin primes between 1 and 100 are; (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73). Hope this helped<3
6 0
3 years ago
Why do we need to use moles when we try to determine amounts of reactants and products in a reaction?.
ICE Princess25 [194]

\huge\fbox{Answer ☘}

<em>Chemists use the mole unit to represent 6.022 × 10 23 things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro's number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units. </em>

hope helpful~

8 0
3 years ago
In a titration, 4.7 g of an acid (HX) requires 32.6 mL of 0.54 M NaOH(aq) for complete reaction. What is the molar mass of the a
katrin2010 [14]

Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

Volume of NaOH = 32.6 ml = 0.0326 L

Molarity of NaOH = 0.54 M = 0.54 mole/L

First we have to calculate the moles of NaOH.

\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole

Now we have to calculate the moles of an acid.

In the titration, the moles of an acid will be equal to the moles of NaOH.

Moles of an acid = Moles of NaOH = 0.017604 mole

Now we have to calculate the molar mass of and acid.

\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}

Now put all the given values in this formula, we get:

0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}

\text{Molar mass of an acid}=266.985g/mole

Therefore, the molar mass of an acid is 266.985 g/mole

3 0
3 years ago
Question 116 g of sunflower oil is
Luba_88 [7]
<h2>Answer:  125.41 mL</h2>

Explanation:

Volume = mass ÷ density

             =  116 g  ÷  0.925 g/mL

             =  125.41 mL

<h3>A 116 g of sunflower oil of 0.925 g/mL has a volume of 125.41 mL.</h3>
7 0
3 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
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