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3 years ago

Help please !!!!!!!!!

Chemistry

1 answer:

Sveta_85 [38]3 years ago

6 0

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

Use dimensional analysis and equation coefficients to convert moles of Al2O3 to moles of O2, to grams of O2:

[(1.00 mol Al2O3)/1][(3 mol O2)/(2 mol Al2O3)][(31.998 g O2)/(1 mol O2)] =

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How do u name this aromatic organic?

Mashutka [201]

A two carbon chain attached to a benzene ring: ethylbenzene

6 0

3 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

The theoretical yield of NaBr from

Lapatulllka [165]

Taking into account definition of percent yield, the percent yield for the reaction is 100%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 FeBr₃ + 3 Na₂S → Fе₂S₃ + 6 NaBr

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

FeBr₃: 2 moles
Na₂S; 3 moles
Fе₂S₃: 1 mole
NaBr: 6 moles

<h3>Moles of NaBr formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 2 moles of FeBr₃ form 6 moles of NaBr, 2.36 moles of FeBr₃ form how many moles of NaBr?

$moles of NaBr=\frac{2.36 moles of FeBr_{3}x6 moles of NaBr }{2 moles of FeBr_{3}}$

moles of NaBr= 7.08 moles

Then, 7.08 moles of NaBr can be produced from 2.36 moles of FeBr₃.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield}x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, being the molar mass of NaBr 102.9 g/mole, you know:

actual yield= 7.08 moles× 102.9 g/mole= 728.532 grams
theorical yield= 7.08 moles× 102.9 g/mole= 728.532 grams

Replacing in the definition of percent yields:

$percent yield=\frac{728.532 grams}{728.532 grams}x100$

Solving:

<u><em>percent yield= 100%</em></u>

Finally, the percent yield for the reaction is 100%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

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8 0

2 years ago

How many atoms are in this formula? 4 O2 A 2

Effectus [21]

Answer:

8 atoms as 4*2=8

Explanation:

4 0

3 years ago

The burning of a sample of propane

r-ruslan [8.4K]

Answer:

70.0°C

Explanation:

We are given;

Amount of heat generated by propane as 104.6 kJ or 104600 Joules
Mass of water is 500 g
Initial temperature as 20.0 ° C

We are required to determine the final temperature of water;

Taking the initial temperature is x°C

We know that the specific heat of water is 4.18 J/g°C

Quantity of heat = Mass × specific heat × change in temperature

In this case;

Change in temp =(x-20)° C

Therefore;

104600 J = 500 g × 4.18 J/g°C × (x-20)

104600 J = 2090x -41800

146400 = 2090 x

x = 70.0479

=70.0 °C

Thus, the final temperature of water is 70.0°C

7 0

3 years ago