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3 years ago

Which of the gases in air are compounds? Explain how you can tell.

Chemistry

2 answers:

Eduardwww [97]3 years ago

5 0

Answer:

$\huge \underline\bold\red{Answer}$

The air in the atmosphere consists of nitrogen, oxygen, which is the life-sustaining substance for animals and humans, carbon dioxide, water vapour and small amounts of other elements (argon, neon, etc.). Higher in the atmosphere air also contains ozone, helium and hydrogen.

Ksivusya [100]3 years ago

3 0

Answer:

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Tiny organisms known as cyanobacteria, or blue-green algae.

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3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

7. Suppose each of these isotopes emits a beta particle. Give the iso-

Volgvan

In a beta emission, the mass number of the daughter nucleus remains unchanged while the atomic number of the daughter nucleus increases by one unit. The following are isotopes produced when the following undergo beta emission;

1) potassium-42 ------> Ca - 42

2) iodine-131 ------------> Xe - 131

3) iron-52 ---------------> Co - 52

4) sodium-24 -----------> Mg -24

The daughter nucleus formed after beta emission is found one place after its parent in the periodic table.

Regarding the stability of the daughter nuclei, a nucleus is unstable if the neutron-proton ratio is less than 1 or greater than 1.5.

Hence, the following daughter nuclei are stable; Ca - 42, Xe - 131, Mg -24.

Learn more: brainly.com/question/1371390

7 0

3 years ago