Data Entry - No Scoring During your procedure to determine the uncalibrated volume near the stopcock of your buret, what was the mass in grams of water that drained out when the meniscus started just above the 25 mL mark and ended just above the stopcock? Report to the thousandth of a gram.

0.025 grams

Step-by-step explanation:

The water in the stopcock has a volume of 25 mL initially, After that, the whole water was drained out. So we have:

Volume of drained water = (25 mL)(1 x 10⁻⁶ m³/1 mL)

Volume of drained water = 25 x 10⁻⁶ m³

Density of drained water = 1000 kg/m³

So, for the mass of drained water:

Density of drained water = Mass of drained water/Volume of drained water

Mass of drained water = (Density of drained water)(Volume of drained water)

Mass of drained water = (1000 kg/m³)(25 x 10⁻⁶ m³)

Mass of drained water = 0.025 gram

Density

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

4.08%

step-by-step explanation: