If you were given two transparent liquids how would you determine which is pure and which is salt water

Why are there more fossils of marine organism than of land organisms o dc

MatroZZZ [7]

Answer:

Three primary reasons. First, there is simply more water-covered places than dry ground places for the animals and plants to have lived. Second, the seas are much more crowded with the kinds of life that leave fossils than the land is. Third, the process that form fossils work very well under water.

Explanation:

4 0

3 years ago

For the reaction N2(g) + 3H2(g) 2NH3(g), what will happen if hydrogen gas was removed from the reaction mixture? There will be a

Hoochie [10]

This uses something called <span>Le Chatelier's principle. It states essentially that any stress put upon a system will be corrected.

In more simple terms, it means that in an equilibrium, such as the equation N2(g) + 3H2(g) <=> 2NH3(g), removing a reactant will cause the system to create more of said reactant to compensate for its loss, or adding excess reactant will cause the system to remove some of the added reactant. For future reference, the same principle applies to products in an equilibrium as well.

In this case, hydrogen gas is a reactant, and hydrogen is being removed. According to </span><span>Le Chatelier's principle, the system will shift to create more hydrogen gas. In essence, it will shift in the direction of the hydrogen gas, so there will be a shift toward the reactants.

To clear something up, Keq will not change, as it is a constant value with constant conditions (such as temperature, pressure, etc.).</span>

7 0

3 years ago

Essentially, the oxygen we breath is

stepan [7]

Answer:

b

Explanation:

Plants need carbon dioxide and let out oxygen

3 0

3 years ago

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

3 years ago

Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when

trapecia [35]

Answer:

$CuCl_{2}(aq.)$ will form a precipitate of insoluble $CuCO_{3}$ when aqueous $Na_{2}CO_{3}$ is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and $NH_{4}^{+}$

all salts of sodium are soluble

When $Na_{2}CO_{3}$ is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of $Na_{2}CO_{3}$ into $CuCl_{2}(aq.)$ will result precipitation of insoluble $CuCO_{3}$

Reaction: $Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)$

4 0

3 years ago