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MAXImum [283]
3 years ago
11

Look at the following diagram of the carbon cycle.

Chemistry
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

The energy consumed by animals in the form of glucose is conserved because it is transformed into chemical energy as carbon dioxide is produced during respiration.

Explanation:

There's no diagram....but I kinda figured it from the description.

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10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Zolol [24]
1 Aluminium is oxidised Al - 3e = Al⁺³

2 Chlorine is reduced  Cl⁺⁷ + 8e = Cl⁻¹

3 Nitrogen is oxidised 2N⁻³ - 6e = N₂

4 0
3 years ago
Andi performs the calculation that is shown below.
rodikova [14]

Answer: 3.59

Explanation:

(2.06)(1.743)(1.00)

2.06 × 1.743 × 1.00

= 3.59058

Two of the multiplied digits are represented in 3 significant figures. Therefore, for correct representation, the result of the product should be written to three significant figures.

3.59058 to 3 significant figures:

First three digits = 3.59

Fourth digit '0' is less than 5, and thus rounded to 0 with other succeeding digits

Therefore, (2.06)(1.743)(1.00) to 3 significant figures equals :

3.59

3 0
3 years ago
Shifts in the rock layer locations cannot account for gaps in the rock record<br><br> true or false
vivado [14]

Answer: That would be false because it is the contact between two layers representing a gap in the geologic record, usually from the erosion of the layers which would normally be expected to appear.

Explanation:

Have a good day

I hope this helps if not sorry :(

Stay motivated

4 0
3 years ago
24) What is momentum? *
sergiy2304 [10]

Answer:

Momentum is the measure of the motion of an object found by multiplying the objects mass and velocity.

Symbol: p

Units: kg x m/s

Explanation:

......

6 0
2 years ago
Iodine-131 decays with a half-life of 8.02
dlinn [17]
Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = \frac{2.303}{t} X log \frac{initial.conc}{final.conc}

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = \frac{2.303}{6.01} X log \frac{5}{final.conc}
∴ final conc. = 2.975 mg
3 0
3 years ago
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