97.22 grams of magnesium is needed.
<u>Explanation:</u>
The molar ratio of Mg to O2 to make MgO is 2:1
Here, we have 2 moles of O2, therefore we need 4 moles of Mg.
1 mol of Mg = 24.305 g
Therefore
4 moles of Mg = 4 
24.305 g
= 97.22 g
Hence we need 97.22grams of magnesium are needed to completely reactwith 2.00 mol of O2 in the synthesis reaction that producesmagnesium oxide.
Answer:
Amylase.
Explanation:
The process of digestion begin to start in mouth when food mix with saliva. An enzyme is released which is called Amylase help in digestion of carbohydrates.
<span>You will use an ideal gas equation here. Let us denote 'x' as the unknown sample of zinc if another 1 ml of it is added to make a total of 476mL gas. The ideal gas equation is PV = nRT. Assume atmospheric conditions and then substitute everything to the equation.
(1atm)(0.476L) = (1.85 + x)(0.08206 L-atm/mol-K)(20.5+273K)
x = 1.83 g of zinc</span>
Answer:
ρ = 4407.03 kg/m³
Explanation:
The Density of a metal alloy is given by the following equation:
1/ρ = m₁/ρ₁ + m₂/ρ₂ + m₃/ρ₃
where,
ρ = density of allow = ?
ρ₁ = density of Titanium (Ti) = 4540 kg/m³
ρ₂ = density of Aluminum (Al) = 2710 kg/m³
ρ₃ = density of Vanadium (V) = 6110 kg/m³
m₁ = mass fraction of Titanium (Ti) = 90% = 0.9
m₂ = mass fraction of Aluminum (Ti) = 6% = 0.06
m₁ = mass fraction of Vanadium (V) = 4% = 0.04
Therefore,
1/ρ = 0.9/(4540 kg/m³) + (0.06)/(2710 kg/m³) + (0.04)/(6110 kg/m³)
1/ρ = 1.9823 x 10⁺⁴ m³/kg + 0.2214 x 10⁻⁴ m³/kg + 0.0654 x 10⁻⁴ m³/kg
1/ρ = 2.2691 x 10⁻⁴ m³/kg
ρ = 1/(2.2691 x 10⁻⁴ m³/kg)
<u>ρ = 4407.03 kg/m³</u>
