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beks73 [17]
3 years ago
14

Which statement is true for most autotrophic organisms?

Chemistry
1 answer:
lara [203]3 years ago
5 0

omg yees

eExplanation:

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Leno4ka [110]
A petroleum-based lubricant used to in farming machinery.
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Global warming facts
Bingel [31]
Cattle contribute to more carbon emissions than all of the worlds transport.
5 0
3 years ago
Choose the correct option for the redox reaction below : 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) 4H2O(l) H+ (oxid
hodyreva [135]

Explanation:

8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

To determine if a substance was reduced or oxidized, the oxidation number at the reactant and product side is compared.

When oxidation occurs, an element loses electrons and its oxidation number increases (becomes more positive). When reduction occurs, an element gains electrons and its oxidation number decreases or is reduced (becomes more negative).

H+ (oxidized or reduced) Answer 1 Choose...

There is an increase in oxidation number from 0 to 1. This means H+ was oxidized.

MnO4- (oxidizing or reducing agent) Answer 2 Choose...

In chemistry, an oxidizing agent is a substance that has the ability to oxidize other substances — in other words to accept their electrons

MnO4- is a strong oxidizing agent.

Fe2+ (oxidizing or reducing agent) Answer 3 Choose...

A reducing agent means it will itself get oxidized. In  this reaction, Fe2+ is the reducing agent.

Fe2+ (oxidized or reduced) Answer 4 Choose...

Oxidized

O in MnO4- (oxidized or reduced) Answer 5 Choose...

There is no change in oxidation number of , so it was neither reduced nor was it oxidized.

Mn in MnO4- (oxidized or reduced) Answer 6 Choose...

There is a decreaase in oxidation number from +7 to +2. This means reduction.

6 0
4 years ago
What should go in the blank?
dem82 [27]

Answer:

³⁸₂₀Ca.

Explanation:

³⁸₁₉K –> __ + ⁰₋₁β

Let ʸₓA represent the unknown.

Thus the equation above can be written as:

³⁸₁₉K –> ʸₓA + ⁰₋₁β

Thus, we can obtain the value of y an x as follow:

38 = y + 0

y = 38

19 = x + (–1)

19 = x – 1

Collect like terms

19 + 1 = x

x = 20

Thus,

ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca

Therefore, the equation is:

³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β

6 0
3 years ago
How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
4 years ago
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