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amm1812
3 years ago
9

Define the term diffusion

Chemistry
2 answers:
maw [93]3 years ago
7 0

Answer:

Spreading, like dispersing.

Explanation:

Varvara68 [4.7K]3 years ago
5 0
Diffusion is the movement of substances from a region of high concentration to a region of low concentration
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If a lab requires each lab group to have 25ml of a solution and it takes 15 grams of CuNO3 to make 1 liter of solution how many
tino4ka555 [31]

We need to do some general algebra here.

We will find that you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

<em>We know that:</em>

  • Each lab group needs 25 ml of solution.
  • it takes 15 g of CuNO₃ to make one L of that solution.
  • There are 22 labs.

Because each lab needs 25 ml of solution, 22 labs will need that amount 22 times, so the <u>total amount of solution needed</u> is:

22*25ml = 550 ml

Now we know that we need 15 grams to make one liter of solution, and:

1 L = 1000ml

Then you need 15g to make 1000ml

and x (we want to find this amount) to make 550ml

Then we can write two equations (not actual equations, as these are different units) like:

x = 550ml

15g = 1000ml

Now we can take the quotient between these two equations:

x/15 g = (550ml/1000ml)

And now we can solve this for x:

x = (550ml/1000ml)*15g = 8.25g

So you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

If you want to learn more, you can read:

brainly.com/question/8743486

5 0
3 years ago
Dosage calculation order: 3 mg available: 2 mg per 6 ml how many ml will be given?
Anna35 [415]

9ml will be given for the case of dosage calculation order: 3 mg available: 2 mg per 6 ml

Conversion factors are necessary for dosage calculation, such as when translating from pounds to kilograms or liters to milliliters. This approach, which is straightforward in design, enables physicians to deal with different units of measurement and convert factors to arrive at the solution.

dosage calculation techniques serve as a second or third check on the accuracy of the previous computation techniques. Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method are the three main approaches for dosage calculation. dosage calculations are frequently prescribed and labeled based on their weight or, for solutions, their strength, which is the amount of weight dissolved or suspended in a given volume.

To learn more about dosage calculation please visit -
brainly.com/question/12720845
#SPJ4

3 0
1 year ago
What is the overall enthalpy of reaction for the equation shown below?
Rudiy27

Answer:

ΔH₁₂ = -867.2 Kj

Explanation:

Find enthalpy for 3H₂ + O₃ => 3H₂O given ...

2H₂ + O₂ => 2H₂O      ΔH₁ = -483.6 Kj

        3O₂ => 2O₃        ΔH₂ = + 284.6 Kj

_____________________________

3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O       (multiply by 3 to cancel O₂)

6H₂ + 3O₂ => 6H₂O        ΔH₁ = 3(-483.6 Kj) = -1450.6Kj

          2O₃ => 3O₂           ΔH₂ = -284.6Kj              (reverse rxn to cancel O₂)

_______________________________

6H₂ + 2O₃ => 6H₂O         ΔH₁₂ = -1735.2 Kj       (Net Reaction - not reduced)

________________________________

divide by 2 => target equation (Net Reaction - reduced)

3H₂ + O₃ => 3H₂O            ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj    

4 0
3 years ago
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate
Vika [28.1K]

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}

7 0
3 years ago
Please need me know! Chem question
Inessa [10]
90 degrees
As temperature increases, rate of reaction increases :)
5 0
3 years ago
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