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3 years ago

What is the percent composition of each of the elements in Al(OH) 3 ?

Chemistry

2 answers:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

%Al = 34.6%

%0 = 61.5%

%H = 3.9%

Explanation:

Al(OH)3

Atomic mass of Aluminum Al = 27

Atomic mass of Oxygen O=16

Atomic mass of Hydrogen H=1

Molar mass of Al(OH)3 = 27 + (16*3) + (1*3)

Al(OH)3 = 27 + 48 + 3

Al(OH)3 = 78

Percentage composition;

Aluminum Al= (27/78) × 100

Aluminum Al=0.346 × 100 =34.6%

Oxygen O=(16*3/78) × 100

Oxygen O= 0.615 × 100 =61.5%

Hydrogen H= (1*3/78) × 100

Hydrogen H=0.039 × 100=3.9%

stellarik [79]3 years ago

3 0

Answer:

Answer Below

Explanation:

Percent composition by element

Element Symbol Mass Percent

Aluminium Al 34.590%

Hydrogen H 3.877%

Oxygen O 61.533%

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What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$

$4.2 + log\frac{0.05}{0.045} = 4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

$pH = pKa + log\frac{base}{acid}$

$4.245 = 3.75 + log\frac{base}{acid}$

$log\frac{base}{acid} = 0.5$

$\frac{base}{acid} = 3.162$

Now we must solve the equation above. This will be done using the following values

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L$

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

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3 0

1 year ago

What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr

Masja [62]

Answer: 116 g of copper

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds = 4.00 hr = $4.00\times 3600s=14400s$ (1hr=3600s)

$Q=24.5A\times 14400s=352800C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500C=193000C$ of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = $\frac{63.5}{193000}\times 352800=116g$ of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus remaining in solution = (0.1-0.003)=0.097moles

8 0

2 years ago

A series of chemicals were added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution to produce a precipi

OLEGan [10]

Answer:

First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of $HNO_{3}$ .

Explanation:

In presence of NaCl, $AgNO_{3}$ forms an insoluble precipitate of AgCl.

Reaction: $Ag^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)$

In presence of $NH_{3}$ , AgCl gets dissolved into solution due to formation of soluble $[Ag(NH_{3})_{2}]^{+}$ complex.

Reaction: $AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)$

In presence of $HNO_{3}$ , $[Ag(NH_{3})_{2}]^{+}$ complex gets destroyed and free $Cl^{-}$ again reacts with free $Ag^{+}$ to produce insoluble AgCl

Reaction: $[Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)$

4 0

2 years ago

How many grams of CH4 will be in a 500ml contain at STP?

ArbitrLikvidat [17]

Answer:

Mass = 0.32 g

Explanation:

Given data:

Mass of CH₄ = ?

Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)

Temperature = 273 K

Pressure = 1 atm

Solution:

Volume of CH₄:

500 mL (500 mL× 1L/1000 mL= 0.5 L)

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

By putting values,

1 atm× 0.5 L = n×0.0821 atm.L/ mol.K × 273 K

0.5 atm.L = n×22.4 atm.L/ mol

n = 0.5 atm.L / 22.4 atm.L/ mol

n = 0.02 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.02 mol × 16 g/mol

Mass = 0.32 g

7 0

2 years ago

At 100.0∘C, the ion product of water is 5.13×10−13. What is the concentration of hydroxide ions at this temperature? Your answer

Trava [24]

Answer: The concentration of hydroxide ions at this temperature is $7.16\times 10^{-7}$

Explanation:

When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.

The ionic product for water is written as:

$K_w=[H^+][OH^-]$

$5.13\times 10^{-13}=[H^+][OH^-]$

As $[H^+]=[OH^-]$

$5.13\times 10^{-13}=x^2$

$x=7.16\times 10^{-7}$

The concentration of hydroxide ions at this temperature is $7.16\times 10^{-7}$

6 0

2 years ago