1. different gases mix together to form air and no new comound is formed
2. air has variable composition
3. air shows the properties of its constituent substances
Answer:
- <u>The pointer will not be lined up with the zero mark; it will be above the zero mark.</u>
Explanation:
Althoug the list of statements is not provided, you can assure that the pointer will not be lined up with the zero mark, but above the zero mark.
A <em>triple beam balance</em> is very precise instrument, of common use in most labs, used to measure masses. It has three beams. Each beam works for different precisions and mass increments. One beam measures 1 g increments, other beam measures 10 g increments, and the other one 100 g increments.
If you place a mass of 250 grams on the measurement tray, then the weight will be balanced in the beams with 200 grams in the beam that reads increments of 100 g and 50 grams in the beam that reads increments of 10 g. When this is the situation, the weight is completely balanced, and the pointer will be lined up with the zero mark. But this is not the case.
When the rider on the 500 gram beam is set to the 200 gram mark, and the other riders are set to 0 grams, the weight of the mass (250 g) on the tray is greater than the force exerted by the rider on the beam (200 g) and so the tray will be below the level of balance, and the beam will be above the level of balance. The level of balance is indicated when the pointer is lined up with the zero mark.
Hello!
Ok so for this problem we use the ideal gas law of PV=nRT and I take it that the scientist needs to store 0.400 moles of gas and not miles.
So if we have
n=0.400mol
V=0.200L
T= 23degC= 273k+23c=296k
R=ideal gas constant= 0.0821 L*atm/mol*k
So now we rearrange equation for pressure(P)
P=nRT/V
P=((0.400mol)*(0.0821 L*atm/mol*k)*(296k))/(0.200L) = 48.6 atm of pressure
Hope this helps you understand the concept and how to solve yourself in the future!! Any questions, please feel free to ask!! Thank you kindly!!!
Plasma, Gas, Water, Solid
25%
Explanation:
In a half life of a radioactive isotope is 1 day,it means it loses its half mass each day
We a formula for N half life
where n is the number of days
Here the isotope is kept for 2 days
so it's left over mass will be
It's left over mass 1/4th of the original mass
Now, we need to find it's percentage by multiplying with 100
<u>So</u><u> </u><u>2</u><u>5</u><u>%</u><u> </u><u>mass</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>2</u><u> </u><u>day</u><u> </u><u>of</u><u> </u><u>half</u><u> </u><u>life</u><u> </u><u>radioactive</u><u> </u><u>isotope</u><u>.</u><u> </u>
