Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)
when we have STP conditions, we can use this conversion: 1 mol = 22.4 L
first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.
molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)
calculations:
Blue
When red litmus paper comes into contact with any alkaline substance, it turns blue. Some examples of alkaline substances are ammonia gas, milk of magnesia, baking soda and limewater.
<span> They </span>do<span> not </span>respond to stimuli<span>, they </span>do<span> not grow, they </span>do<span> not </span>do<span> any of the things we normally associate with life.
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Answer:
Look Below
Explanation:
Capillary action is adhesion of water. In plants, capillary action allows water to move from the roots to leaves. In animals, capillary action plays a role in respiration where oxygen-poor blood oxygenates in the capillaries.
<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>