Answer:
When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.
This decreases the blood glucose level in the level. This is how the negative feedback works in the body if the level of glucose increases.
Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.
Answer:
im not sure but I hope this helps
Explanation:
I believe the equivalents is just the moles reactant/moles limiting reactant
water has a denisty of 1 g/mL. 1 L is 1000 ml so there are 1000g/L.
the molar mass of water is 18g/mol if you use the Liters in the equation above you can find the number of grams present. divide this number you found by 18 to find the moles.
now take the amount of the other reactant given and divide it by its own molar mass. this will give you the number of moles of that reactant.
divide the moles of water by the moles of the reactant and that is the equivalent.
to find the normality you take this number and divide it by the number of liters.
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
so basically
some fuels have an impurity in them which is sulfur.
When the fuel undergoes combustion, the sulfur reacts with oxygen in the air to form sulfur dioxide.
the sulfur dioxide reacts with water vapour in the air to form sulfurous acid, which is a type of acid rain.
Also
the high pressures inside a car engine may cause nitrogen and oxygen in the air to react and form oxides of nitrogen. the most common compounds formed inside car engines are NO (nitrogen oxide) and NO2 (nitrogen dioxide)