1.
Viscosity of honey is higher than that of water.
The water drops falling during rain takes circular shape beacuse of surface tension.
Answer:
There are 4 significant figures
Explanation:
Any number that isn’t 0 is significant. Example:765.Any number even if it’s 0 is significant if it’s after the decimal.example:5.00. I believe if the 0 is at the beginning of the number it’s significant aswell.example:086. 0 isn’t significant if it’s in the middle like this: 306
Answer:
Explanation:
volume of solution = 100 + 100 = 200 mL
mass of reaction mixture = 200 x 1.010 = 202 g
specific heat of solution mixture = 4.184 J/g-°C
heat evolved = mass x specific heat x rise in temperature
= 202 x 4.184 x 7.01
= 5924.63 J
100 mL of 1 M HCl
= .1 L of 1M HCl will contain .1 mole of HCl
.1 mole reacts with .1 mole of NH₄OH to make 5924.63 J of heat
1 mole reacts with 1 mole of NH₄OH to make 59246.3 J of heat
heat of reaction = 59246.3 J / mol
= 59.246 kJ / mol
= 59.3 kJ / mol .
Explanation:
two unpaired electrons...
Answer:
The amount of P₄S₃ produced is 0.16 moles.
Explanation:
The balanced reaction is:
8 P₄ + 3 S₈ → 8 P₄S₃
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate:
- P₄: 8 moles
- S₈: 3 moles
- P₄S₃: 8 moles
Being the molar mass of each compound:
- P₄: 123.895 g/mol
- S₈: 256 g/mol
- P₄S₃: 220.093 g/mol
then by stoichiometry of the reaction, the following amounts of reactant and product participate in the reaction:
- P₄: 8 moles* 123.895 g/mol= 991.16 g
- S₈: 3 moles* 256 g/mol= 768 g
- P₄S₃: 8 moles* 220.093 g/mol= 1,760.744 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
In this case, you calculate the limiting reactant using the following rule of three: if by stoichiometry 768 g of S₈ react with 991.16 g of P₄, 40 g of S₈ with how much mass of P₄ will it react?
mass of P₄= 51.62 grams
But 51.62 grams of P₄ are not available, 20 grams are available. Since you have less mass than you need to react with 40 grams of S₈, P₄ will be the limiting reagent.
Then you can apply the following rule of three: if by stoichiometry 991.16 grams of P₄ form 8 moles of P₄S₃, 20 grams of P₄ will form how many moles of P₄S₃?
moles of P₄S₃= 0.16
<u><em>The amount of P₄S₃ produced is 0.16 moles.</em></u>