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svlad2 [7]
3 years ago
11

Iron is the most

Chemistry
1 answer:
Georgia [21]3 years ago
5 0

hope this is correct

Answer:

D

Explanation:

Iron is the fourth most abundant element, by mass, in the Earth's crust. The core of the Earth is thought to be largely composed of iron with nickel and sulfur. The most common iron-containing ore is haematite, but iron is found widely distributed in other minerals such as magnetite and taconite.

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Calculate the average atomic mas of an unknown element that has two naturally-occurring isotopes.
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The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:

Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
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Write the reaction that occurs when propyl methanoate is hydrolyzed in water and in NaOH
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An ester , propyl methanoate ( HCOOC₃H₇) when reacts with sodium hydroxide( NaOH) forms sodium methanoate (HCOONa) as the main product and propanol (C₃H₇OH).

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HCOOC₃H₇+NaOH ⇒HCOONa + C₃H₇OH

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It is achievable for the same enzyme to catalyze reverse reactions for the reason that the direction of a reversible reaction is determined by the concentrations of reactants and products. In pulmonary circulation, the low CO2 concentration supports the making of CO2 and H2O.
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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
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