Mass of gas produced : 15 g
<h3>Further explanation
</h3>
In general, the reaction takes place in the open air. If the reaction results in the form of gas as in combustion, the mass of the reaction results will be smaller than the original mass.
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
So In a closed/isolated system, the total mass of the substance before the reaction will be equal to the total mass of the reaction product.
Mass of solution A = 70 g
Mass of solution B = 35 g
Mass of mixture : 90 g
Mass of gas : X
Mass of gas X :
Complete Question
A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.
Answer:
The pressure required is 
Explanation:
Generally the volume of a sphere is mathematically denoted as
Substituting 
Converting to Liters
Assume that the pressure at which the air is given to the diver is 1 atm when the air was occupying a volume of 2600L
So
From Charles law
Substituting 
, 
, 
, and making 
the subject we have
Answer:
it's because aluminium is a ductile material
The question is typed wrong. Assuming the correct question is
How many liters of 0.98 M H2SO4 solution would react completely with 3.5 moles Ca(OH)2 ?
Answer:-
3.571 litres
Explanation:-
The balanced chemical equation for this reaction is
H2SO4 + Ca(OH)2 = CaSO4 + 2 H2O
From the balanced chemical equation we see that
1 mole of Ca(OH)2 reacts with 1 mol of H2SO4.
∴3.5 moles of Ca(OH)2 reacts with 1 x 3.5 / 1 = 3.5 mol of H2SO4.
Strength of H2SO4 = 0.98 M
Volume of H2SO4 required = Number of moles of H2SO4 / Strength of H2SO4
= 3.5 moles / 0.98 M
= 3.571 litre
Answer:
7.43 × 10²⁴ m⁻³
Explanation:
Data provided in the question:
Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹
Electron concentration, n = 2.9 × 10²² m⁻³
Electron mobility, 
= 0.14 m²/V-s
Hole mobility, 
= 0.023 m²/V-s
Now,
σ = 
or
σ = 
here,
q is the charge on electron = 1.6 × 10⁻¹⁹ C
p is the hole density
thus,
2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )
or
1.75 × 10²³ = 0.406 × 10²² + 0.023p
or
17.094 × 10²² = 0.023p
or
p = 743.217 × 10²²
or
p = 7.43 × 10²⁴ m⁻³
