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polet [3.4K]
4 years ago
10

What is any condition in an experiment that can be changed

Chemistry
2 answers:
poizon [28]4 years ago
6 0
It would be a dependent variable.
g100num [7]4 years ago
3 0
Dependent variable can be changed in any experiment
You might be interested in
The volume of a diamond is found to be 2.8 ml. what is the mass of the diamond in carats?
Helen [10]

Answer:

49.14 carat.

Explanation:

From the question given above, we obtained the following data:

Volume of diamond = 2.8 mL

Mass of diamond (in carat) =.?

Next, we shall determine the mass of diamond in grams (g). This can be obtained as follow:

Volume of diamond = 2.8 mL

Density of diamond = 3.51 g/mL

Mass of diamond ( in grams) =?

Density = mass /volume

3.51 = mass /2.8

Cross multiply

Mass of diamond (in g) = 3.51 × 2.8

Mass of diamond (in g) = 9.828 g

Finally, we shall Convert 9.828 g to carat. This can be obtained as follow:

1 g = 5 carat

Therefore,

9.828 g = 9.828 g /1 g × 5 carat

9.828 g = 49.14 carat.

Therefore, the mass of the diamond in carat is 49.14 carat.

5 0
3 years ago
An equal number of moles of glucose, C6H12O6, and KI are dissolved in equal volumes of water. Which solution has the higher: (12
Alinara [238K]

Answer:

I dont know the answer

Explanation:

I dont know the answer

3 0
3 years ago
Calculate the amount of heat gained by the water in cup 2 after adding the hot object(s) to it.
Iteru [2.4K]

Answer:

The amount of heat gained by the water in cup 2 after adding the hot object(s) to it is 2119.121 Joules

Explanation:

As we know

Amount of heat gained

Q = mc (T2-T1)

Here,

mass of water in cup 2 (m) = 79.10 grams

Temperature of water in cup 2 = 16.8 degree Celsius

Specific heat of water (c) = 4.186 J/(g °C)

Final Temperature of water in cup 2  = 23.2 degree Celsius

Substituting the given values, we get -

Q = 79.10 * 4.186 * (23.2 -16.8) = 2119.121 Joules

The amount of heat gained by the water in cup 2 after adding the hot object(s) to it is 2119.121 Joules

3 0
3 years ago
What is the molar solubility of aucl3 (ksp = 3. 2 x 10-23) in a 0. 013 m solution of magnesium chloride (a soluble salt)?
Inessa [10]

The molar solubility of AuCl₃ in a 0.013 M solution of magnesium chloride is 1.81×10⁻²⁸M.

<h3>What is Ksp?</h3>

The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution. And for the AuCl₃, Ksp will be written as: Ksp = [Au³⁺][Cl⁻]³

Let the solubility of the AuCl₃ in 0.013M solution of magnesium chloride is x, of Au³⁺ is x and of Cl⁻ is 3x. But we know that MgCl₂ is a strong electrolyte and it completely dissociates into its ions and will produce 2 moles of chloride ions. For this solution let we consider the volume is 1 liter then the concentration of chloride ions in MgCl₂ is 2(0.013)=0.026M.

So, in MgCl₂ solution concentration of Cl⁻ becomes = 3x + 0.026.

Value of Ksp for AuCl₃ = 3.2 × 10⁻²³

On putting all values on the Ksp equation, we get

Ksp = (x)(3x + 0.026)³

Value of 3x is negligible as compared to the 0.026, so the equation becomes

3.2 × 10⁻²³ = (x)(0.026)³

x = 3.2×10⁻²³ / (0.026)³

x = 1.81×10⁻²⁸M

Hence the molar solubility of AuCl₃ in 0.010M MgCl₂ is 1.81×10⁻²⁸M.

To know more about molar solubility, visit the below link:

brainly.com/question/27308068

#SPJ4

4 0
3 years ago
How are some of The physical properties of metals related to the fact that our study supply of electricity is able to reach our
julsineya [31]

I'm pretty sure the answer would be that the metal is conductive, which travels through the copper wires.

7 0
3 years ago
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