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garik1379 [7]
3 years ago
12

These cooked noodles were separated from the water they were cooked in. By which physical property were they separated? A. size

B. density C. magnetism D. boiling point
Chemistry
1 answer:
Papessa [141]3 years ago
4 0

Answer:

B. Density

Explanation:

Density is the only property of liquid in the option that is feasible.

Water has no form of magnetism as it’s not a magnetic substance.

Water carries the size of the container it is put into and has no definite shape.

Water’s boiling point can only be gotten when it’s heated and it’s known to be mostly 100 degree celsius when pure and may be more or less depending on the level of impurities.

However Density is the mass per volume of water and can be calculated before the noodle is cooked which is more effective in its differentiation.

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A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

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

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
3 years ago
List first thirty elements with their valences​
lbvjy [14]

Element   Atomic Number Valency

Valency of Hydrogen 1 1

Valency of Helium 2 0

Valency of Lithium 3 1

Valency of Beryllium 4 2

Valency of Boron 5 3

Valency of Carbon 6 4

Valency of Nitrogen 7 3

Valency of Oxygen 8 2

Valency of Fluorine 9 1

Valency of Neon 10 0

Valency of Sodium (Na) 11 1

Valency of Magnesium (Mg) 12 2

Valency of Aluminium 13 3

Valency of Silicon 14 4

Valency of Phosphorus 15 3

Valency of Sulphur 16 2

Valency of Chlorine 17 1

Valency of Argon 18 0

Valency of Potassium (K) 19 1

Valency of Calcium 20 2

Valency of Scandium 21 3

Valency of Titanium 22 4

Valency of Vanadium 23 5,4

Valency of Chromium 24 2

Valency of Manganese 25 7, 4, 2

Valency of Iron (Fe) 26 2, 3

Valency of Cobalt 27 3, 2

Valency of Nickel 28 2

Valency of Copper (Cu) 29 2, 1

Valency of Zinc 30 2

5 0
3 years ago
I don't know which one it is!
ch4aika [34]
It's what you put :) 
ΔH is the distance from the reactants (which would be E), to the products (which would be G or D)
4 0
3 years ago
How many different d orbits are within the 3d sublevel
ohaa [14]

Answer:

Its 5

Explanation:

6 0
3 years ago
Read 2 more answers
Describe one way to prove that a mixture of sugar and water is a solution and that a mixture of sand and water is not a solution
Llana [10]

Answer:

The solution is always homogeneous mixture and transparent through which the light can travel. The mixture of water and sugar is a solution because sugar is soluble in water and form homogeneous mixture while the sand can not dissolve in water and sand particles scatter the light.

Explanation:

Solution:

"The solution is always homogeneous mixture and transparent through which the light can travel"

The mixture of water and sugar is a solution because sugar is soluble in water and form homogeneous mixture. The solubility of sugar is high as compared to the sand in water because the negative and positive ends of sucrose easily dissolve into the polar solvent i.e, water

Suspension:

"Suspension is the heterogeneous mixture, in which the solute particles settle down but does not dissolve"

The mixture of water and sand is suspension. The sand can not dissolve in water because it is mostly consist of quartz. The nonpolar covalent bonds of  sand are too strong and cannot be break by water molecules.

7 0
3 years ago
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