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Delvig [45]
3 years ago
15

Select the statement about mixtures that is correct. Select one: a. Suspensions are homogeneous mixtures of two or more componen

ts. b. Solutions contain particles that settle out in time. c. Suspensions can change reversibly from liquid to solid. d. A solution contains solvent in large amounts and solute in smaller quantities.
Chemistry
1 answer:
andreev551 [17]3 years ago
5 0

Answer:option d. A solution contains solvent in large amounts and solute in smaller quantities.

Explanation:

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Equation 3x-5=-2x+ 10?<br> O<br> 0<br> x= 5<br> -5 = x<br> -15 = -5x<br> -5x = 15
Harman [31]

Answer:

are those the answer choice

7 0
3 years ago
How many moles of oxygen would be needed to produce 84 moles of sulfur trioxide according to the following balanced chemical equ
olganol [36]

Answer:

126 moles

Explanation:

2S +3 o2=2so3

So if 2 moles of so3 required 3 moles of oxygen

. So 84 moles of so3 will require 84*3/2=126 moles of oxygen

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3 years ago
) What parts of pizzly resemble that of a grizzly bear?
user100 [1]

Answer:

Animal was shot in Nunavut, Canada earlier this month by hunters. It resembles a polar bear but has claws and brown paws of a grizzly bear. Experts claim the 'pizzly' or 'grolar' bear is a hybrid between the two. DNA tests revealed it was actually a blonde haired grizzly bear

Explanation:

6 0
3 years ago
The ksp of yttrium fluoride, yf3, is 8.62 × 10-21. calculate the molar solubility of this compound.
motikmotik

Answer:

The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.

Explanation:

In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.

       YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)

I                       0               0

C                     +S            +3S

E                       S              3S

The solubility product (Ksp) is:

Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴

S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21}  /27}=4.23 \times 10^{-6}M

6 0
3 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0
3 years ago
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