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Archy [21]
2 years ago
6

What is the speed of each vehicle? (Remember S=d/t)

Chemistry
1 answer:
Aleksandr [31]2 years ago
5 0

Answer:

A: 20 km/h, B: 10 km/h

Explanation:

Speed of Vehicle A = 40-20/ 2-1

=20 km/h

If A is 20km/h, B is 10km/h as given in option

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How can substances in a compound and in a mixture be separated?
Ray Of Light [21]
<span>A chemical reaction is required to separate the substances in a compound. The components of a mixture can be separated based on their physical properties using techniques like filtration or distillation.</span>
8 0
3 years ago
A sample of 1.00 moles of oxygen at 50°C and 98.6 kPa, occupies what volume?
enyata [817]

Answer:

27.22 dm³

Explanation:

Given parameters:

number of moles = 1 mole

temperature= 50°C, in K gives 50+ 273 = 323K

Pressure= 98.6kpa in ATM, gives 0.973 ATM

Solution:

Since the unknown is the volume of gas, applying the ideal gas law will be appropriate in solving this problem.

The ideal gas law is mathematically expressed as,

Pv=nRT

where P is the pressure of the gas

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

Input the parameters and solve for V,

0.973 x V = 1 x 0.082 x 323

V= 27.22 dm³

5 0
3 years ago
What does carbon dioxide absorb the most heat energy
Ilia_Sergeevich [38]
Should be <span>during deposition.

</span>
8 0
2 years ago
What example of mixture could be seperated using two or more techniques? Explain your answer
STatiana [176]
You can take two liquids of different densities (how much mass is in a given volume) and pour them into a funnel. An example is oil and water. When the mixture settles, the denser liquid will be at the bottom, and drips through the funnel first. This is a separation that you can just let occur naturally.
8 0
2 years ago
We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim
eduard

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know  

Ku = Vp * Kair + Vmagnesium * K metal  

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit  

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

8 0
2 years ago
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