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bixtya [17]
3 years ago
7

Pick the correct answer down below

Chemistry
2 answers:
never [62]3 years ago
6 0

The correct answer is A I believe!

Tema [17]3 years ago
4 0

the correct answer is d)

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Write the proper word equation to express the following chemical reaction: 4Fe (s) + 3Sn(NO3)4 (aq) arrow 4Fe(NO3)3 (aq) + 3Sn (
Lapatulllka [165]
The word equation for this reaction is:

Iron + Tin nitrate → Iron nitrate + Tin
3 0
3 years ago
Which subshells (s, p, d, f, or g) can have electrons with the indicated magnetic quantum number (ml)?
amm1812

Answer:

=3 means is 3 or greater so that would be f and g subshells

=0 means is 0 or greater so that would be s, p, d, f and g subshells

=1 means is 1 or greater so that would be p, d, f, and g subshells

=4 means is 4 or greater so that would be g only

4 0
2 years ago
How many grams of Pb are contained in a mixture of 0.135 kg each of PbCl(OH) and Pb2Cl2CO3?
Dahasolnce [82]

Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg)  = 135 g

Molecular mass of PbCl(OH)  = 207+35.5+16+1  = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g  PbCl(OH)  :

(207 g Pb /  259.5 g PbClOH) * 135g PbClOH  =

0.79768 * 135 =>  107.68 g of Pb

For Pb2Cl2CO3  :

Given the mass of Pb2Cl2CO3  :

0.135 Kg = 0.135 Kgx(1000g / 1Kg)  = 135 g

Molecular mass of Pb2Cl2CO3  = 2*207+2*35.5+12+3*16  = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3  = 2*207 = 414 g

Hence mass of Pb in 135 g  Pb2Cl2CO3:

(414 g Pb /  545 g PbClOH) * 135g PbClOH  =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3


Hope that helps!

8 0
3 years ago
Read 2 more answers
fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she
enot [183]

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

 (0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

7 0
3 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
2 years ago
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