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3 years ago

What makes the factor that makes the sky appear blue?

Chemistry

1 answer:

Charra [1.4K]3 years ago

4 0

Answer:

Gases and particles in Earth's atmosphere scatter sunlight in all directions. Blue light is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

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Mass of 60.009 g and radius of 1.02 cm what is the density of the sphere in (g/ml)

alexandr402 [8]

Density is given as mass / volume.

Mass is the sphere is 100 g.

Volume of the sphere = (pi∗r3)∗4/3
(
p
i
∗
r
3
)
∗
4
/
3

=(4∗22∗3∗3∗3)/(7∗3)cm3
=
(
4
∗
22
∗
3
∗
3
∗
3
)
/
(
7
∗
3
)
c
m
3

=792/7
=
792
/
7
cm3
3

Therefore, Density is 100/(792/7)g/cm3
100
/
(
792
/
7
)
g
/
c
m
3

Which gives: density = 0.883838 g/cm3
g
/
c
m
3

If you want to change the units to kg per cubic metres, then we need to divide this value by 1000( for g to kg) and multiply by 100 * 100 * 100 (for cm to m).

This makes the density to be 883.83 kg/m3

4 0

3 years ago

What is true about the molar mass of chlorine gas? the molar mass is 35.5 g. the molar mass is 71.0 g. the molar mass is equal t

Bingel [31]

Chlorine gas, or dichlorine, is composed of 2 chlorine atoms and has a molar mass of 70.96, or about 71 g. So the answer is the molar mass is 71 g.

Hope this helps!

6 0

3 years ago

Assuming you start with 10.0 g of zinc and 10.0 g hydrochloric acid, identify the limiting reagent and determine what mass of th

Liula [17]

The balanced chemical reaction for the substances given would be as follows:

Zn + 2HCl = ZnCl2 + H2

We are given the amounts of the reactants used in the reaction. We use these amounts to determine which is the limiting and excess reactant. We do as follows:

10 g Zn (1 mol / 65.38 g) = 0.1530 mol
10 g HCl (1 mol / 36.46 g) = 0.2743 mol

From the the stoichiometric ratio which is 1 is to 2, the limiting reactant would be hydrochloric acid and the excess would be zinc metal.

Mass of zinc that remains = 0.1530 - (0.2743 / 2) = 0.0159 g Zn

5 0

3 years ago

A beaker contains 175.32g of salt, NaCl(58.44g/mol). Water is added until the final volume is 2 liters. The solution should be d

cricket20 [7]

Answer:

E. 1.5M

Explanation:

because there is total 175.32 g of NaCl dissolved in 2 liter of water, and 58.44 g = 1 mol

so total miles = 175.32/58.32 = 3 moles

now 3 moles are in 2 liter of water, hence one liter contains 3/2= 1.5

Therefore answer is E. 1.5M

8 0

3 years ago

Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com

lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30 = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0

3 years ago