A sample of an ideal gas at 1.00 atm and a volume of 1.73 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 70.0 atm, what was the volume of the sample

Volume is 0.0247L

Explanation:

This question involves the use of Boyle's law which states that the volume of a given mass of gas is inversely proportional to it's pressure provided that temperature remains constant.

Mathematically,

v = k/p

K = VP

P1V1 = P2V2 = P3V3 = =PnVn

V = volume

p = pressure

Data;

P1 =1.0atm

V1 = 1.73L

P2 = 70atm

V2 = ?

P1V1 = P2V2

V2 = P1V1 / P2

V2 = (1.0×1.73)/70

V2 = 0.0247L

The volume of the sample is 0.0247L

b. acid is the correct answer @jayxboof all social media

acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol