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3 years ago

How is air temperature usually measured?

Chemistry

2 answers:

andrew11 [14]3 years ago

8 0

Thermometer is thermometer

wel3 years ago

3 0

thermometer

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8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

$T^2\propto R^3$

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

$(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3$

$(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3$

$\dfrac{1}{64}=(\dfrac{R}{R'})^3$

$R'=4\times R$

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.

4 0

3 years ago

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:

Step 1: 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

Step 2: 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

Step 3: 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

Step 4: 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.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: 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

All of the following are true of carbohydrates except

Evgen [1.6K]

3. they make up larger compounds like fats and amino acids, would be the correct answer.

5 0

3 years ago

Read 2 more answers

If the reactants contain one chlorine molecule, do you know how many chlorine atoms will be in the product? Include conversion o

max2010maxim [7]

Answer:

The number of Chlorine atoms in the product is 2.

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction.

The reactants contain one chlorine molecule( $Cl_{2}$ ) which has two chlorine atoms.

Then, according to law of conservation of matter, the product must contain two chlorine atoms.

4 0

3 years ago

Which graph shows a negative acceleration ( position and time )

allsm [11]

I'm pretty sure its graph 4

7 0

3 years ago