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3 years ago

What would happen to temperatures on earth the suns heat were distributed throughout the atmosphere

2 answers:

8 0

If the sun's heat is distributed uniformly throughout the atmosphere, this will ensure that each point on the earth gets a moderate level of heat and temperature in these locations will be within the acceptable level.
If the sun's heat is concentrated in one part of the atmosphere, the location that correspond to the point of concentration will receive enormous amount of heat from the sun, which will shoot up the temperature of that location. This will leads to all manner of disruptions in the ecosystem that is found in that location.

wlad13 [49]3 years ago

8 0

For example, California's water cycle was on hold because we did not have any rain. The weather has been dry warm, so all of the water is being evaporated, but never reaching the precipitation phase of the cycle.

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Carbon-14 (14C) dating assumes that the carbon dioxide on the Earth today has the same radioactive content as it did centuries a

Nataly [62]

Answer: The tree was burned 16846.4 years ago to make the ancient charcoal

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5715 years

Putting values in above equation, we get:

$k=\frac{0.693}{5715yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = 13 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{13}\\\\t=16864.4yrs$

Hence, the tree was burned 16846.4 years ago to make the ancient charcoal

8 0

3 years ago

What happens when the pressure of a gas is decreased?

Serggg [28]

Answer:

The combined gas law states that the pressure of a gas is inversely related to the volume and directly related to the temperature. If temperature is held constant, the equation is reduced to Boyle's law. Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase.

Explanation:

3 0

3 years ago

Elements with atomic numbers 7 and 83 have how many valence electrons? *

IRINA_888 [86]

Explanation:

element with atomic numbers 7 and 83 have 5 and 5 valence electrons

5 0

3 years ago

A researcher claims that an ancient scroll originated from greek scholars in about 500 bce. a measure of its carbon-14 decay rat

Naddik [55]

The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,

A(t) = (A(o))(0.5)^(t/h)

where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,

0.89 = (0.5)(t / 5730 years)

The value of t from the equation is 963.34 years.

Answer: 963 years

8 0

4 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago