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Dominik [7]
3 years ago
15

What is a colligative property?

Chemistry
1 answer:
V125BC [204]3 years ago
6 0
<span>Colligative properties are properties of solutions that depend on the number of molecules [or ions] in a given volume of solvent and not on the properties (e.g. size or mass) of the compound. Colligative properties include: lowering of vapor pressure; elevation of boiling point; depression of freezing point and osmotic pressure.</span>
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Using complete sentences, explain how to predict the products and balance the reaction for the combustion of butane, C4H10. C4H1
hichkok12 [17]
At the complete combustion of butane the carbon dioxide and water are formed.

2C₄H₁₀ + 13O₂ = 8CO₂ + 10H₂O

C₄H₁₀ ⇒ 4CO₂, 5H₂O ⇒ 13O ⇒ 6.5O₂
we use integer coefficients: 8CO₂, 10H₂O, 13O₂
3 0
3 years ago
A rigid container of O has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K?
tia_tia [17]

Answer:

P₂ = 130.18 kPa

Explanation:

In this case, we need to apply the Gay-Lussack's law assuming that the volume of the container remains constant. If that's the case, then:

P₁/T₁ = P₂/T₂   (1)

From here, we can solve for the Pressure at 273 K:

P₂ = P₁ * T₂ / T₁   (2)

Now, all we need to do is replace the given data and solve for P₂:

P₂ = 340 * 273 / 713

<h2>P₂ = 130.18 kPa</h2>

Hope this helps

4 0
3 years ago
What is the minimum amount of 6.0 M H2SO4H2SO4 necessary to produce 25.0 g of H2(g)H2(g) according to the reaction between alumi
Lelechka [254]

2.083 Liters of 6.0 M solution sulfuric acid is required. This solved using molecular calculations and Titration.

Solution: 2Al(s)+3H_2SO_4(aq) = Al_2(SO_4)_3(aq)+3H_2(g)

Moles of hydrogen gas =  \frac{25}{2} = 12.5 mol

Then 12.5 moles of hydrogen will be obtained from Moles of Sulfuric acid = 12.5 mol

Molarity of the sulfuric acid solution = 6.0 M = 6 mol/ l

6M = \frac{12.5 mole}{V}

where V is the volume needed

V = \frac{12.5}{6}

V = 2.083 l

<h3>What is Titration?</h3>
  • Titration, commonly referred to as titrimetry, is a typical quantitative chemical analysis method used in laboratories to ascertain the unidentified quantity of an analyte .
  • Titration is frequently referred to as volumetric analysis because it relies heavily on volume measurements. The titrant or titrator is a reagent that is prepared as a standard solution.
  • To determine concentration, a solution of the analyte or titrand reacts with a known concentration and volume of the titrant. The titration volume is the amount of titrant that has responded.
  • Titrations come in a variety of forms with various protocols and objectives. Redox and acid-base titrations are the two most typical types of qualitative titrations.

To learn more about titration with the given link

brainly.com/question/2728613

#SPJ4

8 0
2 years ago
What is an Atom? please help
JulsSmile [24]

Answer:

An atom is the basic building block of matter. Anything that has a mass-- in other words, anything that occupies space--is composed of atoms.

5 0
3 years ago
Read 2 more answers
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur
lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

q=m\times c\times \Delta T

q = heat absorbed by water

m = mass of water = 40.0 g

T_{final} = final temperature of water = 20.0°C

T_{initial = initial temperature of water = 17.0°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

q=40.0\times 4.186\times (20.0-17.0)]

q=502J

Hence, the joules of heat were re-leased by the lead is 502

5 0
3 years ago
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