<h3>
<u>Answer;</u></h3>
= 930.23 mL
<h3><u>Explanation</u>;</h3>
Using the combined gas law;
P1V1/T1 = P2V2/T2
Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and
V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K
Thus;
V2 = P1V1T2/T1P2
= (600 ×800 ×500) / (258 × 1000)
= 930.23 mL
Answer:
1.d = 2.70 g/mL
2.d = 13.6 g/mL
4.d = 1896 g / 212.52 cm3 = 8.9 g/cm3
Explanation:If this helped subscribe to Amiredagoat Yt
2Mn⁺⁴O2 + 2K2CO3 + O2 → 2KMn⁺⁷O4 + 2CO2
Mn⁺⁴ - 3e⁻ = Mn⁺⁷
Given:
Atomic radius of gold = 144 x 10⁻¹² m = 144 pm
Density of gold = 19.3 g/cm³ = 19.3 x 10⁻³ kg/cm³
Mass of gold sample = 1.40 g = 1.4 x 10⁻³ kg
Calculate the volume of an atom.
v = [(4π)/3]*(144 x 10⁻¹² m)³ = 1.2508 x 10⁻²⁹ m³
Calculate the mass of an atom.

Calculate the number of atoms in a sample of mass 1.40 g.

Answer: 5.8 x 10²¹ atoms
- 402 has 3 sig figs.
- 0.00560 has 3 sig figs.
- 34.20 has 3 sig figs.
- 6546 has 4 sig figs.
- 45000 has 2 sig figs.
hope this helps :)