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puteri [66]
3 years ago
5

What is a spelling bee​

Chemistry
2 answers:
Lisa [10]3 years ago
8 0

Answer: a contest in which you are eliminated if you fail to spell a word correctly.

mart [117]3 years ago
4 0
A contest where you attempt to spell a series of words that you are given by the judges
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A balloon contains 2.0 L of air at 101.5 kPa. You squeeze the balloon to a volume of .75 L. What is the pressure of air on the i
Nataly_w [17]

You would use the formula for Boyle's Law:

(P1) (V1) = (P2) (V2)

(101.5) (2.0) = (P2?) (.75)

*P2 = 270kPa (You're allowed 2 significant figures)

P = Pressure

V = Volume

3 0
2 years ago
Read 2 more answers
How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
lisov135 [29]
Moles of glucose = Molarity x volume solution 
                              = 4.5 x 1.5
                              = 6.75 moles.

Hope this helps, have a great day ahead!
3 0
3 years ago
Help me
Valentin [98]

Answer:

South American

Explanation:

When you look at a map of plates, only South American forms a boundary with the African plate out of those specific plates

6 0
3 years ago
Read 2 more answers
what is the  concentration of a NaCl solution, in Molarity, if you add 59.76 g of NaCl into 270 mL of H2O
enot [183]

Hey there!

Molar mass NaCl = 58.44 g/mol

Number of moles

n =  mass of solute / molar mass

n = 59.76 / 58.44

n = 1.0225 moles of NaCl

Volume in liters:

270 mL / 1000 => 0.27 L

Therefore:

M = number of moles / volume ( L )

M = 1.0225 / 0.27

= 3.78 M

Hope that helps!

7 0
3 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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