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3 years ago

How does matter move and change through photosynthesis??

1 answer:

3 0

The inputs of photosynthesis are light energy, and matter in the form of water absorbed through the roots, and carbon dioxide absorbed through the leaves. ... The main outputs are oxygen, which is released into the air, and glucose sugar (chemical energy), which is used to keep the plant alive.

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Cl2(g) + 2kBr(s) ---> 2KCl(s) + Br2(g) Rewrite the equation and write the color of each chemical under its name. 2. What type

kolezko [41]

Answer:

Cl2(g) (green/yellow mix) + 2KBr(s) (white) ---> 2KCl(s) (violet) + Br2(g) (reddish brown)

This chemical reaction is a redox type.

Explanation:

Look at the oxidation state, when the number increase your element gets oxidated, when the number decrease, the elements it's getting reduced.

8 0

3 years ago

An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron?

creativ13 [48]

Answer:

D. 450 J/kgK

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where:

Q = amount of heat absorbed/released (J)

m = mass of substance (g)

c = specific heat capacity

∆T = change in temperature (°C)

According to the information provided in this question:

Q = 267.3 kJ = 267300J

m = 18kg

∆T = 318K - 285K = 33K

c = ?

Q = m × c × ∆T

c = Q ÷ m∆T

c = 267300 ÷ 18 × 33

c = 267300 ÷ 594

c = 450 J/kgK

7 0

3 years ago

Watching the way clouds move in the sky would most likely provide helpful information on which characteristic of the weather?

scoundrel [369]

Answer:

By watching the movement of cirrus clouds you can tell from which direction weather is approaching. When you see cirrus clouds, it usually indicates that a change in the weather will occur within 24 hours. Cirrostratus clouds are thin, sheetlike high clouds that often cover the entire sky.

Explanation:

im smart;)

4 0

3 years ago

An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in jou

nevsk [136]

Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = $\frac{Q}{T}$

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{293}$

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

= 12.5 J/K

7 0

3 years ago

Can someone please tell me this answer

ra1l [238]

Answer:

Carbon + oxygen

C + O2 - CO2

Sulphur + Oxygen

S + O2 - SO2

5 0

3 years ago