If a 3 L balloon at 2 atm was compressed to 1 L. What is
the new pressure of the balloon?

6atm

Explanation:

Using Boyle's law equation as follows;

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

Based on the information provided in this question;

P1 = 2 atm

P2 = ?

V1 = 3L

V2 = 1L

Using P1V1 = P2V2

2 × 3 = P2× 1

6 = P2

P2 = 6atm

The new pressure of the balloon is 6 atm.

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer