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3 years ago

Identify the most oxidized compound. Group of answer choices CH3CH2CHO CH3CH2CH3 CH3CH2CH2OH CH3CH2OCH3 CH3CH2COOH

Chemistry

1 answer:

Setler79 [48]3 years ago

7 0

Answer:

Huh!?

Explanation:

explain me please

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How many grams of Co are needed to react with an excess of fe203 to produce 156.2 g FE? show your work.

Gekata [30.6K]

The reaction is given as

Fe2O3 (s)+ 3CO(g)--->3CO2(g)+ 2Fe(s)

No.of moles=mass in gram/molar mass

As for Fe mole =156.2g/55.847=2.7969~2.797

The ratio b/w CO and Fe is 3:2

Moles of CO needed= 2.797x3/2=4.1955

Mass of CO needed= 4.195mol x 28.01g/mol= 117.515g

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3 years ago

How many total atoms are in 0.290 g of P2O5?

Sliva [168]

There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert grams of P₂O₅ to moles of P₂O₅.

$\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\$

Step 2. Convert moles of P₂O₅ to molecules of P₂O₅.

$\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\$

$= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\$

Step 3. Convert molecules of P₂O₅ to atoms.

There are seven atoms in 1 mol P₂O₅.

∴ $\text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\$

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3 years ago

On a mission to a newly discovered planet, an astronaut finds gallium abundances of 61.29 % for 69Ga and 39.71 % for 71Ga. What

snow_lady [41]

I believe it is 70.281amu

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3 years ago

Which equation is derived from the combined gas law?

Aloiza [94]

Answer:

The correct answer is V1/T1=V2/T2.

Explanation:

Just took the test

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3 years ago

Read 2 more answers

In the combustion chamber of an engine, the initial volume is 450 cm3 at a pressure of 1.0 atmosphere. If the piston displaces 4

Alona [7]

Assuming that the contents of the chamber ar ideal gases. We can use the relation PV=nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = (1)(450)/ 48

P2 = 9.375 atm

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3 years ago