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givi [52]
3 years ago
12

A solution that contains a large amount of solute would be described as what

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
8 0

Answer:

A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.

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Help me plz Identify the state of matter.
musickatia [10]
Solids liquids and more
8 0
3 years ago
Read 2 more answers
When two molecules of methanol (CH3OH) react with oxygen, they combine with three O2 molecules to form two CO2 molecules and fou
Alex17521 [72]

Answer:

188

Explanation:

For every 2 molecules of methanol reacted, 4 molecules of water are formed.  Use this relationship to solve.

2/4 = 94/x

2x = 376

x = 188

188 molecules of water will be formed.

3 0
3 years ago
A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an
professor190 [17]

Answer:

1.717 J/g °C  ( third option)

Explanation:

A piece of the unknown metal dropped into water this means that Q of metal is equal to Q of the water. We write this equality as follows:

<u>Step 1: writing the formulas:</u>

Q = mc∆T

⇒ -Q(metal) = Q(water)   Because :Metal dropped into water this means that Q of metal is equal to Q of the water.

<em>We can write the formula different :</em>

Mass of metal * (cmetal)(ΔT) = Mass of water *(cwater) (ΔT)

⇒ Here c is the specific heat and depends on material and phase

<em>For this case :</em>

mass of the metal  = 68.6g

mass of the water = 42g

Specific heat of the metal = TO BE DETERMINED

Specific heat of the water = 4.184J/g °C

Initial temperature of the metal = 100 °C  ⇒ Change of temperature: 52.1 - 100

Initial temperature of the water = 20°C  ⇒ Change of temperature:  52.1 - 20

<u />

<u>Step 2: Calculating specific heat of the metal</u>

-(Mass of metal) * (cmetal)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-68.6g (cmetal)(52.1 - 100) = 42g (4.184j/g °C) (52.1 - 20)

-68.6g *cmetal * (-47.9) = 42g (4.184j/g °C) *(32.1)

3285.94 * cmetal = 5640.87

cmetal = 5640.87 / 3285.94 = 1,71667 J/g °C

cMetal = 1.717 J/g °C

4 0
3 years ago
If there is currently 50kg of U-235 present in Oklo, how much must have been present 750 million years ago when the reaction too
Viefleur [7K]
To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.

The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:

\frac{750}{703.8}  = 1.066
Now, take this scale factor and multiply it by the current mass, like this:

50 \times 1.066 = 53.3
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:

50 + 53.3 = 103.3
I hope this helped you. Fell free to ask any further questions.
7 0
3 years ago
In witch of these examples does chemical energy change to electrical energy
Anuta_ua [19.1K]
D the chemical energy in a batery changes to electrical when its used
7 0
3 years ago
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