Density is equal to mass divided by volume.
So D=118/12
D=9.83
Shielding effect being attendency of the inner electrons to repel the outermost electrons, an increase in shielding effect increases atomic radius and decrease reduces the atomic radius
1.55 g/L; 9.503 03 × 10^20 mL
<em>Part 1.
</em>
<em>Step 1</em>. Convert <em>kilograms to grams</em>.
1.55 kg × (1000 g/1 kg) = 1550 g
<em>Step 2</em>. Convert <em>cubic metres to litres</em>.
1 m^3 × (1000 L/1 m^3) = 1000 L
<em>Step 3</em>. Divide <em>grams by litres</em>
1.55 kg/1 m^3 = 1550 g/1000 L = 1.55 g/L
<em>Part 2.</em>
<em>Step 1</em>. Convert <em>cubic kilometres to cubic metres</em>
950 303 km^3 × (1000 m/1 km)^3 = 9.503 03 × 10^14 m^3
<em>Step 2</em>. Convert <em>cubic metres to litres</em>.
9.503 03 × 10^14 m^3 × (1000 L/1 m^3) = 9.503 03 × 10^17 L
<em>Step 3</em>. Convert <em>litres to millilitres</em>.
9.503 03 × 10^17 L × (1000 mL/1 mL) = 9.503 03 × 10^20 mL
Answer:
Beginning or Start
Explanation:
Reactions can be followed by measuring changes in concentration, mass and volume of reactants or products. The rate is highest at the start of the reaction because the concentration of reactants is highest at this point.
Hope this helped! :^)
= 6.022 × 1020
Explanation<em>;</em>
Mole of aluminium oxide (Al2O3) is
⇒ 2 x 27 + 3 x 16
Mole of aluminium oxide = 102 g
i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.051 molecules)
= 3.011 x 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)
= 2 × 3.011 × 1020
=<em> 6.022 × 1020</em>
<em>hope </em><em>it </em><em>helps</em><em>_</em>
