Equation 3x-5=-2x+ 10? O 0 x= 5 -5 = x -15 = -5x -5x = 15

How would a solution that is labeled 5.0 M would be read as?

Ivanshal [37]

5 moles because molarity signifies number of moles dissolved in One litre of water

8 0

4 years ago

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All scientists try to base their conclusions on _____.

crimeas [40]

They try to base their conclusions off of data and measurements of which they should record from conducting experiments

4 0

3 years ago

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Help me now plssssssss i will give brainliest!!!!!!!!!!!!!!!!!!!!!!!

Blababa [14]

Products are copper+ aluminium chloride
reactants are aluminium+copper chloride

5 0

3 years ago

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

Semenov [28]

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0

3 years ago

Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g

damaskus [11]

Answer:

156.4g K

Explanation:

I'm not sure if it is correct but I think it should be this

What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2

What are we looking for?: #g of K

What is the ratio of K to Cl2?: 2:1

Set up equation: 2molCl2 x $\frac{2mol K}{1 mol Cl2}$

Cancel unwanted units: 2 x $\frac{2mol K}{1}$

Answer we got: 2 x 2mol K = 4mol K

Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K

3 0

3 years ago