Which statement describes absolute and apparent brightness?

Suggest a reason why the noble gases are referred to as being in Group 0 rather than Group 8.

34kurt

Answer:

The atoms of noble gases already have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why the noble gases are inert and do not take part in chemical reactions. The table summarises the electronic configurations of elements in groups 1, 7 and 0.

Explanation:

The atoms of noble gases already have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why the noble gases are inert and do not take part in chemical reactions. The table summarises the electronic configurations of elements in groups 1, 7 and 0.

8 0

3 years ago

Which of the following is NOT a skill scientists use to learn about the world? a. Predicting c. Inferring b. Observing d. Scient

omeli [17]

You good though mamas got to Scientific and it was growing in general

6 0

2 years ago

Consider a 12.01 g sample of carbon. the average mass of a carbon atom is 1.994 3 10223 g. how many carbon atoms are in the samp

Nonamiya [84]

the mass of a carbon atom is 1.994 x 10⁻²³ g
the mass of the carbon sample is 12.01 g
to find the number of Carbon atoms we have to divide the mass of sample by mass of a carbon atom.
number of C atoms = $\frac{12.01 g}{1.994* 10^{-23}g }$
therefore number of atoms = 6.023 X 10²³ atoms of carbon

3 0

4 years ago

What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M<br> solution?

love history [14]

Answer:

Approximately $19.1\; \rm g$ .

Explanation:

<h3>Number of moles of formula units of magnesium sulfate required to make the solution</h3>

The unit of concentration in this question is " $\rm M$ ". That's equivalent to " $\rm mol \cdot L^{-1}$ " (moles per liter.) In other words:

$c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}$ .

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

$\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L$ .

Calculate the number of moles of $\rm MgSO_4$ formula units required to make this solution:

$\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}$ .

<h3>Mass of magnesium sulfate in the solution</h3>

Look up the relative atomic mass data of $\rm Mg$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm Mg$ : $24.305$ .
$\rm S$ : $\rm 32.06$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm MgSO_4$ using these values:

$M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}$ .

Using this formula mass, calculate the mass of that (approximately) $0.159\; \rm mol$ of $\rm MgSO_4$ formula units:

$\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}$ .

Therefore, the mass of $\rm MgSO_4$ required to make this solution would be approximately $19.1\; \rm g$ .

3 0

3 years ago

When you pour 125 grams of solute into 1.25 Liters of 35 grams of crystal sink to the bottom and do not dissolve. What is the sa

Sati [7]

Answer:

72 g/L

Explanation:

The dissolved amount of solute is the difference between the amount you have poured and the amount that precipitated:

125 g - 35 g = 90 g

Thus, 90 grams of solute were dissolved in 1.25 liters. The saturation point is the ratio between the grams dissolved and the volume in liters:

saturation point = 90 g/1.25 L = 72 g/L

4 0

3 years ago