Answer:
calcium ions are released from its store inside the sarcoplasmic reticulum, into the sarcoplasm (muscle ).
Explanation:
KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which <span>then will react with 4-nitroaniline to form a diazonium salt.</span>
Answer:
a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance
Explanation:
Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.
<span>pm stands for picometer and picometers are units which can be used to measure really tiny distances. One picometer is equal to 10^{-12} meters. We know that one centimeter is equal to 10^{-2} m so there are 10^2 cm per meter.
We can change the distance d = 115 pm to units of centimeters.
d = (115 pm) x (10^{-12}m / pm) x (10^2 cm / m)
d = 115 x 10^{-10} cm = 1.15 x 10^{-8} cm
The distance in centimeters is 1.15 x 10^{-8} cm</span>
Answer: 9.9 grams
Explanation:
To calculate the moles, we use the equation:

a) moles of 

b) moles of 


According to stoichiometry :
1 mole of
combine with 1 mole of
Thus 0.33 mole of
will combine with =
mole of
Thus
is the limiting reagent as it limits the formation of product.
As 1 mole of
give = 1 mole of 
Thus 0.33 moles of
give =
of 
Mass of 
Thus theoretical yield (g) of
produced by the reaction is 9.9 grams