Answer:
PUT THE TOOTHPAST ON CABBAGE AND CHECK IT OUT AFTER A DAY
Answer:
I. Increasing pressure will allow more frequent successful collision between particles due to the particles being closer together.
II. Rate of reaction increases due to more products being made; as increased pressure favours the exothermic side of the equilibrium.
III. Increasing temperature provides particles lots of (Kinetic) energy, for more frequent successful collision due to the particles moving at a faster rate than before. However, favouring the endothermic side of the equilibrium due to lots of energy required to break and form new bonds.
IV. Rate of reaction increases due to increase temperature favouring both directions of the equilibrium - causing products to form faster.
Hope this helps!
Answer: 150 kPa
Explanation:
Given that,
Original volume of gas V1 = 30L
Original pressure of gas P1 = 105 kPa
New pressure of gas P2 = ?
New volume of gas V2 = 21L
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
105 kPa x 30L = P2 x 21L
3150 kPa L = P2 x 21L
P2 = 3150 kPa L / 21 L
P2 = 150 kPa
Thus, 150 kPa of pressure is required to compress the gas
Answer:
Explanation:
Answer:
A. to determine the efficiency of the reaction
Explanation:
Percentage is the ratio of the actual yield to theoretical yield as a percentage. It is calculated by dividing the actual yield by theoretical yield then multiplying by 100%.
Calculation of percentage yield is important as it helps in the determination of efficiency of a reaction. For example in most industries for the purpose of making the most product with the least waste.
Additionally, calculating the percentage yield helps in determining other products that may be formed during the reactions.
Answer:
"HEMOGLOBINA. Las características de la hemoglobina (Hb) como amortiguador están íntimamente vinculadas a la capacidad de disociación del grupo imidazólico del aminoácido histidina unido al hierro que contiene el grupo hemo (ver figura inferior)."
Link: http://www.ehu.eus/biomoleculas/buffers/buffer4.htm#:~:text=HEMOGLOBINA&text=Las%20caracter%C3%ADsticas%20de%20la%20hemoglobina,hemo%20(ver%20figura%20inferior).
