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DerKrebs [107]
2 years ago
5

If the ratio of the length of a rectangle to its width is 3 to 2, what is the

Chemistry
1 answer:
borishaifa [10]2 years ago
8 0

Answer:

Length of a rectangle whose width is 4 inches is

<u>6 inches</u>

Explanation:

Ratio of length  to its width is 3 to 2

Let length  = L

Let Width  = B

L: B = 3 : 2 (given)

\frac{L}{B} =\frac{3}{2}

Width ,B = 4

Insert  B in above equation

\frac{L}{B} =\frac{3}{2}

\frac{L}{4} =\frac{3}{2}

cross multiply,

L = \frac{3}{2}\times4

L = 6 inches

Conversion(if required) ,

1 inch = 2.54 cm

6 inch = 2.54 (6)

L = 15.24 cm

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Answer:

2.335 J/g*degrees celcius

Explanation:

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3 years ago
The solubility KI is 50 g in 100 g of H2O at 20 °C. if 110 grams of ki are added to 200 grams of H2O ________
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The solubility KI is 50 g in 100 g of H₂O at 20 °C. if 110 grams of ki are added to 200 grams of H₂O <u>the </u><u>solution </u><u>will be </u><u>saturated</u><u>.</u>

<h3>What is solubility?</h3>

Solubility is a condition where the solute is fully dissolved in the solvent. When fully mixed with the solvent.

Given that 50 g of KI is added to 100 g of water at 20 °C it means 100 g of water can dissolve a maximum of 50 g of  KCl.

1 g of water will dissolve an quantity of 0.5 g of  KCl.

To assay for 200 g of water: 200 g of water can disintegrate a maximum of (0.5) x 200 g of  KCl.

The maximum amount of KCl that will dissolve is 100 g

Actualised amount dissolved = 110 g

when Amount dissolved > Maximum solubility limit

110 g > 100 g

Thus,  the solution is saturated.

To learn more about solubility, refer to the below link:

brainly.com/question/8591226

#SPJ4

6 0
1 year ago
The weak type of bonding which attracts molecules to molecules is:
SashulF [63]
B. Hydrogen is the answer
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3 years ago
How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

4 0
2 years ago
Read 2 more answers
How much liquid does this graduated cylinder contain?
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Answer:

I think 44 mL

Explanation:

I hope this helps u :D

8 0
3 years ago
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