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2 years ago

What metalloid has commonly been used as an insecticide due to its effectiveness as a poison.

Chemistry

1 answer:

Taya2010 [7]2 years ago

5 0

Answer:

Arsenic.

Explanation:

Hello there!

In this case, since insecticides are substances that act as poisons to get rid of insects in order to prevent their presence and/or reproduction in houses, companies, crops and others, a substance that has been widely used is the metalloid arsenic due to its direct affection of the insect's body (movement, performance, cellular functions).

In addition, high levels of arsenic in food could cause arsenic poisoning in humans as well, that is why such practice must be properly performed and by using the correct security protocol.

Best regards!

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Mn2+(aq) + nabio3(s) → bi3+(aq) + mno4−(aq) + na+(aq) how many water molecules are there in the balanced equation (for the react

stiks02 [169]

<span>
Mn²⁺ + 4H2O -----> MnO4⁻ + 8H⁺ +5e⁻ /*2
<span>NaBiO3 +6H⁺ +2e⁻ -----> Bi³⁺ + Na⁺ + 3H2O /*5
</span>2Mn²⁺ + 5 NaBiO3+8H2O+30H⁺ ---> 2MnO4⁻ +5Bi³⁺ + 5Na⁺ +16H⁺ +15H2O

</span>2Mn²⁺ + 5 NaBiO3+14H⁺ ---> 2MnO4⁻ +5Bi³⁺ + 5Na⁺ +7H2O

There are 7 water molecules in this reaction.

4 0

2 years ago

Read 2 more answers

PLEASE HELP URGENT WILL GOVE BRANLIEST WITH EXPLANATION

Ymorist [56]

Answer:

no because it already had it in there and it exploded by a fuse

Explanation:

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6 0

2 years ago

How did oxygen come up to make 21% of the earth's present atmosphere?

MakcuM [25]

Tiny organisms known as cyanobacteria, or blue-green algae.

4 0

2 years ago

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

2 years ago

Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc

docker41 [41]

Answer:

53.6 grams of silver chloride was produced.

Explanation:

$AgNO_3+HCl+\rightarrow AgCl+HNO_3$

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

Mass of silver chloride + Mass of nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

$x=50.0 g+50.0 g - 46.4 g = 53.6 g$

53.6 grams of silver chloride was produced.

8 0

2 years ago