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EleoNora [17]
3 years ago
8

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

Chemistry
1 answer:
Irina18 [472]3 years ago
5 0

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

All you have to do is to create a ratio between the molecule and the oxygen atom.

5 moles of Mg3(PO4)2 (4x2 moles O/1 mole Mg3(PO4)2) = 40 moles of oxygen

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Periodic Table: Why are group 1A elements called alkali metals?
FrozenT [24]

Answer:

Explanation:

The elements in Group I of the periodic table are called alkali metals. They are called alkali metals because they react with water to form alkali solutions. These metals are very reactive; hence they have to be stored under oil to protect them from corrosion by air and waterwaterwater

5 0
2 years ago
As of right now, which of the following forms of nuclear power is the most feasible for human use?
nikitadnepr [17]

As of now, the nuclear fission is the most feasible energy source for human use. All the nuclear power plants are based on the controlled nuclear fission reaction, where the unstable nucleus is bombarded with high speed neutrons, thus, splitting the nucleus into stable ones and releasing huge amount of energy. The nuclear fusion requires very high temperature, the temperature equal's to that of the sun. Hence, it is not feasible right now. As the technology advances, we will see advancement in other form of energies.

6 0
3 years ago
Read 2 more answers
What volume of Co2 (carbon (iv) oxide)
hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0
4 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
Katarina [22]

Answer:

empirical formula: H_2SO_4

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: H_2SO_4.

7 0
3 years ago
BRAINLIST
garri49 [273]
A) divide by 100. A meter is 100 centimeters so that’s how you can tell
3 0
3 years ago
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