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3 years ago

How many significant figures are in the measurement, 0.005890 g?

1 answer:

6 0

4 significant figures. Use a rule called the Atlantic-Pacific rule where if the period is absent (Atlantic), you would start counting the numbers from right to left. If the period is present (Pacific), then start counting the numbers from left to right. Since the period is present, start from 5 and there are 4 digits including 5.

You might be interested in

What pH value can be assigned to acids and bases, respectively?

goldfiish [28.3K]

Acids have a pH less than 7 (pH < 7)
Bases have a pH more than 7 (pH > 7)
A pH of 7 would be neutral

Hope this helped!

5 0

3 years ago

Given that the density of water is 1.00 g/mL, the density of toluene is 0.867 g/mL, and the density of chloroform is 1.48 g/mL,

12345 [234]

Answer:
Top-Toluene
Middle-Water
Bottom-Chloroform

Explication:
Chloroform is on the bottom layer because it is the densest liquid.
Toluene is on the top because it is the least dense.
Water is between the two because it’s density is between chloroform and toluene.

4 0

3 years ago

The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to

adelina 88 [10]

Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:

MW 44 5x32 4x18 3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.

6 0

4 years ago

Human bone does not contain?

tia_tia [17]

Human bone does not contain oxygen

4 0

3 years ago

A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh

Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0

3 years ago