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2 years ago

Phosphorous is prepared according to the following equation: Ca3(PO4)2 + 3SiO2 + 5C = 3CaSiO3 +5CO + 2P

Chemistry

1 answer:

ivann1987 [24]2 years ago

6 0

Answer:

48546.7kg or 48546745g

Explanation:

so from this u can work out the amount of moles in phosphorous by doing mass / mr you have to convert the mass in kg to g so you times it by 1000. then divide it by 31 which is the mr of phosphorous. then u can use the molar ratio which is 2:1 . Then use the equation mass= moles*mr

so 97000*1000/31=312903moles

2:1

so 312903/2=156451 moles of ca3(PO4)2

so mass= moles*mr

156451*310.3=48494g

hope this helps u to understand(*°▽°*)

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yulyashka [42]

Answer:

The smallest constituent of an element is termed as an atom. According to Jaiden, an atom comprises two subatomic particles and three prime parts, however, this is incorrect as an atom comprises three subatomic particles and two main parts. The three subatomic particles are the protons, electrons, and neutrons. All these parts are smaller in comparison to an atom, thus, they are considered as subatomic particles.

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3 years ago

Please help!!! will give brainliest!!! 30 points! (need to put roman numeral in answer)

mariarad [96]

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7. Chromium (III) Phosphide

8. Iron (II) Fluoride

9. Gold (III) Chloride

I hope this helps. I'm more than 100% sure that all the answers except for number 7 are correct. I knew all of them off the top of my head except for this one. I hope the other answer has the correct answer for that one. Good luck and have a great day.

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2 years ago

Which organism would have the most variation in the DNA of its offspring?

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The organism that would have the most variation in the DNA of its offspring is the cat (Option C). Meiosis is a type of cell division that generates more genetic variability than asexual types of reproduction.

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Animals (e.g., cats) generate gametes by meiosis which fuse during fertilization to produce new offspring.

Both amoeba and bacteria reproduce by a type of asexual reproduction called binary fission. Moreover, yeasts also reproduce asexually by a process called budding and fission.

Both asexual and sexual types of reproduction generate genetic variability by the emergence of new mutations in daughter cells.

Meiosis generates much more genetic variability than asexual types of reproduction due to two different processes:

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2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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3 years ago

The wheels of a car are able to spin due to the combustion of gasoline inside the car's engine. what type of energy transfer bes

REY [17]

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2 years ago