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3 years ago

When would a life scientist study a nonliving thing such as a rock or lake

Chemistry

1 answer:

VladimirAG [237]3 years ago

3 0

He would study it so he could see how it interacted with the living things around it.

You might be interested in

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

How much heat, in joules, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature

svp [43]

Answer:

50,849.25 Joules

Explanation:

The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:

Q = mcΔT, where ΔT represents the change in temperature.

In the case of the iron block:

m = 75 g

c = 0.449 J/g °C

ΔT = 1535 - 25 = 1510 °C

Therefore,

Q = 75 g x 0.449 J/g °C x 1510 °C

= 50,849.25 Joules

<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>

5 0

3 years ago

Why did the Mars Climate Orbiter crash?

IRISSAK [1]

Answer:

<h3> The Mars Climate Orbiter crashed, because of a navigation error, and a futile to convert english units to metric units.</h3>

4 0

3 years ago

How many grams of carbon dioxide are produced from 0.98 mol of Fe3O4?

nexus9112 [7]

Hello

the answer is 43.129310000000004

Have a nice day

4 0