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NISA [10]
3 years ago
11

What is the oxidation number of each phosphorus atom in H4P2O7

Chemistry
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

+5

Explanation:

The oxidation number of phosphorus can be obtained as follows:

H4P2O7 = 0

4(+1) + 2P + 7(—2) = 0

4 + 2P —14 = 0

Collect like terms

2P = 14 — 4

2P = 10

Divide both side by 2

P = 10/2

P = +5

The oxidation number of phosphorus is +5

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What volume of chlorine gas (Cl2), measured at STP, can be produced by the decomposition of 73.0g of hydrogen chloride gas (HCl)
gayaneshka [121]

Answer:

Explanation:

2HCl    =   H₂   +   Cl₂

2 mole     1 mole      1 mole

73 gram HCl = 73 / 36.5 = 2 mole of HCl

2 moles of HCl will produce 1 mole of chlorine gas .

At STP , one mole of chlorine gas has volume equal to 22.4 litre .

8 0
3 years ago
The human body on average contains 6 liters of blood. If 20 drops are equal to 1milliliter, how many drops of blood are in the a
garri49 [273]

Answer:

120000drops

Explanation:

Average blood in human blood = 6L - 6*1000 = 6000ml

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5 0
3 years ago
A 100-w light bulb is about 16% efficient. how many joules of heat energy are siddipated by the bulb each second
pickupchik [31]

Answer:

84 Joules are wasted per second.

Explanation:

An efficiency of 16% means 84% is wasted as heat energy.  

Therefore;

For a 100W bulb;

Heat wasted = 84/100 × 100 W

                      = 84 W

Therefore;

84 W or 84 joules are wasted as heat per second

3 0
3 years ago
What is the name of the following chemical compound?
Lynna [10]

Answer: Dinitrogen pentoxide

Explanation:

5 0
3 years ago
Read 2 more answers
Copper(II) fluoride contains 37.42% F by mass. Calculate the mass of fluorine (in g) in 55.5 g of copper(II) fluoride.
Mrac [35]

Answer:

There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride

Explanation:

x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen

37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.

So, 100 g of copper (II) fluoride contains 37.42 g of F

55.5 g of copper (II) fluoride contains \frac{37.42\times 55.5}{100}g of F or 20.8 g of F

Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.

5 0
3 years ago
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