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Kazeer [188]
3 years ago
12

PLEASE HELP

Chemistry
2 answers:
Virty [35]3 years ago
8 0

Answer:

I think the answer is D.coral reef, neritic zone

Explanation:

Alborosie3 years ago
8 0

Answer:

d

Explanation:

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I don't know how to solve this​
Misha Larkins [42]

For the first part, use the question M=mol/vol (liters)

To do this, you have the given 1.6 M solution

divide the 360g by the molar mass of ethanol (44.07) to get moles

360/44.07=8.16 mol

so

1.6M = 8.16 mol/x vol

volume: 5.1 Liters

8 0
3 years ago
234+34.1 add and round to 2 significant figures
marin [14]
Like this? 234+34.1= 268.1 then round. If it is less than 5 then you round down if it is more then you round up. Because it is less the final number would be 268.1=268
3 0
3 years ago
A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
VMariaS [17]

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

cw × mw × ΔTw = -cs × ms × ΔTs

(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g

3 0
3 years ago
What is related to the energy of motion of the particles of matter?
Amanda [17]
The Kinitec Molecular Theory of Matter 
Yes Kninetic energy
5 0
3 years ago
what grade is learning What amount of thermal energy (heat) in joules required to raise the temperature of 25 grams of water fro
blsea [12.9K]

Answer:

10425 J are required

Explanation:

assuming that the water is entirely at liquid state at the beginning , the amount required is

Q= m*c*(T final - T initial)

where

m= mass of water = 25 g

T final = final temperature of water = 100°C

T initial= initial temperature of water = 0°C

c= specific heat capacities of water = 1 cal /g°C= 4.186 J/g°C ( we assume that is constant during the entire temperature range)

Q= heat required

therefore

Q= m*c*(T final - T initial)= 25 g * 4.186 J/g°C * (100°C- 0°C) = 10425 J

thus 10425 J are required

7 0
3 years ago
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