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3 years ago

How many atoms of oxygen are in 0.100 mol of silicon dioxide?

1 answer:

3 0

1) Silicon dioxide formula: SiO2 ....... 2 is a subscript for the O atom

2) From the formula you have 1 molecula of SiO2 contains 1 atom of SiO2

3) Then, 0.100 mol of SiO2 contains 0.1 mol of Si

4) Multiply by Avogadro's number: 0.100mol * 6.022*10^23 atoms/mol= 6.02*10^22 atoms

Answer: 6.02*10^22 atoms

You might be interested in

If a plant produces 8.46 mol C6H12O6, how many moles of H2O are needed?

Ierofanga [76]

Answer:

50.76 mol H2O.

Explanation:

The photosynthesis follows the equation:

6CO2 + 6H2O ---> C6H12O6 + 6O2

This means that 6 mol of H2O are needed to obtain 1 mol of C6H12O6 (see the numbers that precedes every molecule to know how many mols are in game).

So we can say that:

1 mol C6H12O6 --------- 6 mol H2O

8.46 mol C6H12O6 -----x= 8.46 x 6 : 1 = 50.76 mol H20

8 0

3 years ago

identify the part of the slow carbon cycle in whitch the total amount of carbon is most likely decreasing and explain why this d

Westkost [7]

Answer:

i just had gthe sme question looked it up and got it WRONG

Explanation:

sorry i ont have a clue

3 0

3 years ago

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

Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo

siniylev [52]

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

3 0

3 years ago

In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1

weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0

3 years ago