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irakobra [83]
3 years ago
7

How many elements are in this chemical formula? 2(NH4)2SO4

Chemistry
1 answer:
sesenic [268]3 years ago
6 0

Answer: 30

Explanation:

Sulfur         S  1  

Oxygen         O  4  

Nitrogen N         2  

Hydrogen H         8

15 in total. For 2 moles, 15*2 = 30

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Polly did an experiment with marbles in a glass bowl to show the movement of particles in solids, liquids, and gases. The experimental set-up is shown below: A glass bowl is shown with four marbles inside it. ... Add water to the bowl so that the marbles start sliding past one another

Explanation:

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A goup of cells that work together to perform a specific function.
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Cell Differentiation and Tissue.

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How many grams of HF are needed to react with 3.0 moles of Sn?
Flauer [41]

Answer:

120g

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction.

Sn + 2HF —> SnF2 + H2

Step 2:

Determination of the number of mole HF needed to react with 3 moles of Sn.

From the balanced equation above,

1 mole of Sn and reacted with 2 moles of HF.

Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.

Step 3:

Conversion of 6 moles of HF to grams.

Number of mole HF = 6 moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn.

3 0
3 years ago
How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose
katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

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

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

6 0
2 years ago
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