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2 years ago

How many dozen (dz) eggs are needed to make 12 muffins? What about 15.5

Chemistry

2 answers:

NARA [144]2 years ago

8 0

Answer:

I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer

romanna [79]2 years ago

6 0

Should be 1.2 I divided and got 1.2

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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]

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

<u>Explanation:</u>

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

2 years ago

Consider the arrangement of gases shown below. If the value between the gases is opened and the temperature is held constant, de

goblinko [34]

Answer:

I don't know what to say . just for points

7 0

2 years ago

What does our state have that few share?

MissTica

Answer:

which state?

Explanation:

7 0

3 years ago

Round off each number to the indicated number of significant figures (sf)

weeeeeb [17]

Answer:

The answer is B

Explanation:

3 0

2 years ago

Read 2 more answers

A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol

Troyanec [42]

Answer : The correct answer for molarity = 0.19 $\frac{mol}{L}$ and Molality = 0.18 $\frac{mol}{Kg}$ .

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

$Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}$ \

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

$Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}$

Molar mass of aniline (C₆H₅NH₂) = 93.13 $\frac{g}{1 mol}$

Mass of aniline = 6.2 g

Plugging values in mole formula :

$Mole = \frac{6.2 g}{93.13 \frac{g}{1 mol}}$

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

$Volume of solution = \frac{350 mL}{1000mL} * 1 L$

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

$Molarity = \frac{0.0665 mol}{0.350 L }$

Molarity = 0.19 M or 0.19 $\frac{mol}{L}$

------------------------------------------------------------------------------------------

2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

$Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}$

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

$Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}$

Plugging value in density formula :

$1.04 \frac{g}{mL} = \frac{ mass }{350 mL}$

Multiplying both side by 350 mL

$1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL$

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

$Mass of solvent = \frac{364 g}{1000g} * 1 Kg$

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

$Molality = \frac{0.0665 mol}{0.364 Kg}$

Molality = 0.18 m or 0.18 $\frac{mol}{Kg}$

3 0

2 years ago