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2 years ago

How does latitude affect climate? Blast blast blast blast blast blast blast

Chemistry

1 answer:

cestrela7 [59]2 years ago

8 0

Answer:

THANKS

Explanation:

There is a relationship between latitude and temperature around the world, as temperatures are typically warmer approaching the Equator and cooler approaching the Poles. There are variations, though, as other factors such as elevation, ocean currents, and precipitation affect climate patterns.

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The burning of propane gas can be represented as a balanced chemical reaction as follows: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) Calcula

Snezhnost [94]

Answer: 20L of H2O

Explanation:

C3H8 + 5O2 → 3CO2 + 4H2O

Recall 1mole of a gas contains 22.4L at stp

5moles of O2 contains = 5 x 22.4 = 112L

4moles of H2O contains = 4 x 22.4 = 89.6L

From the equation,

112L of O2 produced 89.6L H2O

There for 25L of O2 will produce XL of H2O i.e

XL of H2O = (25 x 89.6)/112 = 20L

6 0

3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

3 0

3 years ago

The total pressure of a confined mixture of gases is the sum of the pressures of each of the gases in the mixture. This is known

olganol [36]

Answer:

Explanation:

It is also known as the Dalton’s law of partial pressure. Given a confinement that contains a mixture of gases which do not mix, the total pressure equals the sum of the individual pressures.

The term, which do not mix is necessary because, if the gases are the type that mix, the law will no longer hold as they would have given up their individual identities and hence their individual partial pressure cannot be use to access them anymore.

Hence, the law helps to sum the totality of the pressures of a number of gases which exists together in a confinement and they do not mix. Say we have 3 gases A, B and C. The total pressure is the sum of pressure A, pressure B and pressure C.

5 0

3 years ago

How many moles of oxygen would be required to produce 5 mol of water? (Remember sig figs)

Vikki [24]

Answer:

Since moles are just "some number of particles", and since one O2 molecule contains enough oxygen for two H2O molecules, you need 2.5 moles of O2 for 5 moles of H2O.

7 0

3 years ago

Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,

vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

$N=n\times N_A$

Where : N = Number of atoms

n = moles

$N_A=6.022\times 10^{23}$ = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol × 44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol × 8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol

Mass percentage of gold :

$\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=33.35 \%$

Mass percentage of silver:

$\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=62.80\%$

Mass percentage of copper :

$\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=3.85\%$

7 0

3 years ago