At STP 32 g of O₂ would occupy by the same volume as 4 g of He
<h3>Further explanation</h3>
Complete question
At STP 32 g of O₂ would occupy by the same volume as:
- 4.0 g of He
- 8.0 g of CH₄
- 64 g of H₂
- 32 g of SO₂
Standard Conditions
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So the gas will have the same volume if the number of moles is the same
mol of 32 grams of O₂ :
<em>So mol of 4 g He = mol of 32 g O₂</em>
Your answer is mass and speed! Please give me brainlist :)
Particles in a liquid move freely and liquids have a definite volume and indefinite shape since it occupies the shape of the container in which the liquid is in.
The energy required to heat 40g of water from -7 c to 108 c is
1541000 joules
calculation
Q(heat)= M( mass) x c(specific heat capacity) xdelta t( change in temperature)
M= 40g= 40/1000= 0.04 Kg
C= 335,000 j/kg/c
delta T ( 108 --7= 115 c)
Q is therefore = 0.04 g x 335000 j/kg/c x 115 c = 1541,000 joules
According to what is known about chemical equilibrium and Le Chatelier's principle, when you increase the amount of the reactants, the reaction will be moved to the products, this is because, the most reactants we have the most products we can produce.
From the given choices, the one that goes according to this reason is the third one: The volume of water vapor increases.
