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3 years ago

Explain how temperature and precipitation are affected by latitude and the sun's rays.

2 answers:

8 0

Latitude and Temperature

At the equator, the Sun's rays are most direct. This is where temperatures are highest. At higher latitudes, the Sun's rays are less direct. The farther an area is from the equator, the lower its temperature.

user100 [1]3 years ago

4 0

The suns rays throw heat down which makes the temperature change

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The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

Is lithium ionization energy high low or medium?

satela [25.4K]

It’s soft which makes It low energy

6 0

3 years ago

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What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa

Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

T1 = 30 Degrees or 273.15 + 30 = 303.15 K

P1 = 91 KPa

V2 =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

$\frac{P1VI}{T1}$ = $\frac{P2V2}{T2}$

putting the values in the equation of Gas Law:

V2 = $\frac{P1V1T2}{P2T1}$

V2 = $\frac{91 X 75 X 273.15}{303.15 X 101.3}$

V2 = 60.7 ml

at STP the volume of carbon dioxide gas is 60.7 ml.

4 0

4 years ago

BRAINLIEST ASAP

Sveta_85 [38]

The first three quantum numbers for electrons located in subshell 4s are :

n = 4

l = 0

m = 0

3 0

3 years ago

24 POINTS!!!!!!!!!!

Oliga [24]

Your answer is going to be A, because it was shoved harder, it will go faster

8 0

3 years ago

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