Answer:
One mole of cadmium (6multiply1023 atoms) has a mass of 112 grams, as shown in the periodic table on the inside front cover of the textbook. The density of cadmium is 8.65 grams/cm3.
Explanation:
Answer:
1.6g/mL
Explanation:
Density equation is D=m/v
Density = g/mL
m=mass of sample in grams
v = volume of sample in mL
The volume of a square can be calculated by V=l*w*h.
In this case it is 5cm*5cm*5cm = 125cm^3
Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:
125cm^3 (1mL/(1cm^3)) = 125mL
Then simply plug into the density equation:
D=200g/125mL = 1.6g/mL
Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Answer:
C₄H₉O₂
Explanation:
just count the amount of atoms present in the model.
Answer:
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Explanation:
The HI donates a proton to the water, converting it to a hydronium ion
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Thus, the HI is behaving like a Brønsted acid.
