All categories

2 years ago

How many moles of Si are in 78.0 grams of silicon

Chemistry

2 answers:

Igoryamba2 years ago

8 0

Answer:

$\boxed {\boxed {\sf 2.78 \ mol \ Si}}}$

Explanation:

To convert from grams to moles, we must use the molar mass. This number can be found on the Periodic Table.

Si (Silicon): 28.085 g/mol

We can use number as a ratio.

$\frac { 28.085 \ g \ Si}{ 1 \ mol \ Si}$

Multiply by the given number of grams.

$78.0 \ g \ Si *\frac { 28.085 \ g \ Si}{ 1 \ mol \ Si}$

Flip the fraction so the grams of silicon cancel.

$78.0 \ g \ Si *\frac {1 \ mol \ Si }{ 28.085 \ g \ Si}$

$78.0 *\frac {1 \ mol \ Si }{ 28.085 }$

$\frac {78.0 \ mol \ Si }{ 28.085 }= 2.77728324729 \ mol \ Si$

The original number of grams (78.0) has 3 significant figures, so our answer must have the same. For the number of moles calculated, that is the hundredth place. The 7 in the thousandth place tells us to round the 7 to a 8.

$2.78 \ mol \ Si$

There are about 2.78 moles of silicon in 78.0 grams.

gladu [14]2 years ago

6 0

The answer is 2.78 moles of Si

You might be interested in

When a mixture of iron powder and sulphur is heated it glows more brightly than that of iron filings and sulphur.Explain this ob

Bond [772]

Answer:

I don't know

Explanation:

mixture of iron powder and sulphur heated it glows because sulphur and iron are more reactive metals that they shine and sulphur can react with water produce fire it's write

3 0

2 years ago

This table lists some characteristics of two planets.<br> Which statement is true?

LekaFEV [45]

gravity

Explanation:

gravity is the same in any planet

5 0

2 years ago

Read 2 more answers

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing tin

mojhsa [17]

The half reactions as they occur at each electrode

is as follows

at the anode Sn(s) =sn^2+(aq) + 2e -
at the cathode 2 ag^+(aq) + 2e - = 2Ag (s)

net cell reaction = Sn (s) + 2Ag^+(aq) = sn^2+ (aq) + 2 Ag (s)

8 0

3 years ago

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<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 HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago

Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what

Drupady [299]

Answer:

Explanation:

6 0

3 years ago