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3 years ago

I need help with Number 63

Chemistry

1 answer:

Lubov Fominskaja [6]3 years ago

8 0

Isotopes are variants of an element which have differing number of neutrons. So, yes, because all of them have the same atomic number, meaning they are all Mg, however they all have differing atomic masses meaning different number of neutrons.

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Substance X has a fixed volume, and the attraction between its particles is strong. Substance Y has widely spread out particles

enot [183]

Among the choices given, the correct answer is the first option. Substance X most likely is a crystal, and substance Y most likely is a liquid. Substance X having a fixed volume describes a crystal because crystals occupy a certain volume. Substance Y is a liquid because liquids can still be compressed further in order to attain a more packed structure.

8 0

3 years ago

Read 2 more answers

I need help with this assignment

yuradex [85]

The answers for the following sums is given below.

1. $Pd_{2}H_{2}$

2. $C_{2} H_{6}$

3. $C_{2} H_{2} O_{2} Cl_{2}$

4. $C_{3}Cl_{3} N_{3}$

5. $Tl_{2 } C_{4} H_{4}O_{6}$

6. $C_{8}H_{8}$

7. $N_{2}O_{5}$

8. $P_{4}O_{6}$

9. $C_{4}H_{8} O_{2}$

Explanation:

1.Given:

Molar mass=216.8g

Molecular formula=Pd $H_{2}$

we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of Pd $H_{2}$ :

Pd=106.4×1=106.4u

H=1×2=2

molar mass of Pd $H_{2}$ =106.4+2=108.4

n= $\frac{216.8}{106.4}$

n=2

Molecular formula=2(Pd $H_{2}$ )

Molecular formula= $Pd_{2}H_{2}$

Therefore the molecular formula of the compound is $Pd_{2}H_{2}$

2. Given:

Molar mass=30.0g

Molecular formula= $CH_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of $CH_{3}$ :

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of $CH_{3}$ =12.01 + 3 =15.01u

n= $\frac{30.0}{15.01}$

n=2

Molecular formula=2( $CH_{3}$ )

Molecular formula= $C_{2} H_{6}$

Therefore the molecular formula of the compound is $C_{2} H_{6}$

3. Given:

Molar mass=129g

Molecular formula=CHOCl

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n= $\frac{129}{64.5}$

n=2

Molecular formula=2(CHOCl)

Molecular formula= $C_{2} H_{2} O_{2} Cl_{2}$

Therefore the molecular formula of the compound is $C_{2} H_{2} O_{2} Cl_{2}$

5. Given:

Molar mass=577g

Molecular formula= $TlC_{2} H_{2}O_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of $TlC_{2} H_{2}O_{3}$ :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of $TlC_{2} H_{2}O_{3}$ =204.3+24.02+1+48.00=278.32u

n= $\frac{577}{278.32}$

n=2

Molecular formula=2 ( $TlC_{2} H_{2}O_{3}$ )

Molecular formula= $Tl_{2 } C_{4} H_{4}O_{6}$

Therefore the molecular formula of the compound is $Tl_{2 } C_{4} H_{4}O_{6}$

4. Molar mass=184.5g

Molecular formula=CClN

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n= $\frac{184.5}{61.5}$

n=3

Molecular formula=3 (CClN)

Molecular formula= $C_{3}Cl_{3} N_{3}$

Therefore the molecular formula of the compound is $C_{3}Cl_{3} N_{3}$

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula= $C_{8}H_{8}$

Molecular formula = $C_{8}H_{8}$

Molar mass of $C_{8}H_{8}$ :

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of $C_{8}H_{8}$ =96.08+8=104.08u

n=104.08÷78.0

n=1

Molecular formula = n(Empirical formula)

Molecular formula = 1( $C_{8}H_{8}$ )

Molecular formula = $C_{8}H_{8}$

Therefore the molecular formula of a compound is $C_{8}H_{8}$

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen= $\frac{4.04}{14.01}$ = 0.289 moles

moles of oxygen= $\frac{11.46}{15.999}$ =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula of $N_{2} O{5}$ .

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is $N_{2}O_{5}$ .

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula= $P_{2} O_{3}$

molecular formula= 2(Empirical formula)

Molecular formula =2( $P_{2} O_{3}$ )

Molecular formula = $P_{4}O_{6}$

Therefore the molecular formula of the compound is $P_{4}O_{6}$

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula = $C_{2} H_{4} O$

Molecular formula = 2(Empirical formula)

Molecular formula =2( $C_{2} H_{4} O$ )

Molecular formula = $C_{4}H_{8} O_{2}$

Therefore the molecular formula of the compound is $C_{4}H_{8} O_{2}$

4 0

3 years ago

Can someone please help

oksian1 [2.3K]

Answer:

Even ur best Friend Can you to have red flag feeling

6 0

3 years ago

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac

Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0

3 years ago

A cylinder with a movable piston originally has a volume of 2805 mL and is filled with nitrogen to a pressure of 4.00

Sergio039 [100]

This problem is providing the initial volume and pressure of nitrogen in a piston-cylinder system and asks for the final pressure it will have when the volume increases. At the end, the answer turns out to be 2.90 atm.

<h3>Boyle's law</h3>

In chemistry, gas laws are used so as to understand the volume-pressure-temperature-moles behavior in ideal gases and relate different pairs of variables.

In this case, we focus on the Boyle's law as an inversely proportional relationship between both pressure and volume at constant both temperature and moles:

$P_1V_1=P_2V_2$

Thus, we solve for the final pressure by dividing both sides by V2:

$P_2=\frac{P_1V_1}{V_2}$

Hence, we plug in both the initial pressure and volume and final volume in order to calculate the final pressure:

$P_2=\frac{2805mL*4.00atm}{3864mL}\\ \\P_2=2.90atm$

Learn more about ideal gases: brainly.com/question/8711877

6 0

3 years ago