(B. 3) 172 All nonzero digits are significant.
(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.
(A. 4) 3427 All nonzero digits are significant.
(B. 3) 0.0000455 Leading zeroes are not significant.
(B. 3) 0.00456 Leading zeroes are not significant.
(C. 5) 2205.2 Zeroes between nonzero digits are significant.
(C. 5) 107.20 Trailing zeroes after the decimal point are significant.
(B. 3) 0.0473 Leading zeroes are not significant.
Answer:
It is better to do chemistry
Explanation:
So that you will learn more of chemicals
This question includes four answer choices:
A. definite volume, highest molecular motion, highest kinetic energy
B. indefinite volume, least molecular motion, highest kinetic energy
C. definite volume, least molecular motion, lowest kinetic energy
D. definite volume, no molecular motion, lowest kinetic energy
Solids do not have the highest molecular motion (on the contrary they have the least molecular motion), so you can discard option A. Solids have a definite volume and the highest kinetic energy (given that they have the least molecular motion), so you discard option C. Molecules always have a vibrational motion, so you discard option D. Option C, have only characteristics that correctly describes a solid: definite volume, least molecular motion, lowest kinetic energy. Therefore, the answer is the option C.
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</span>
Answer:
2.038 seconds.
Explanation:
So, in the question above we are given the following parameters in order to solve this question. We are given a rate constant of 0.500 s^-, initial concentration= 0.860 M and final concentration= 0.310 M,the time,t =??.
Assuming that the equation for the first order of reaction is given below,that is;
A ---------------------------------> products.
Recall the formula below;
B= B° e^-kt.
Therefore, e^-kt = B/B°.
-kt = ln B/B°.
kt= ln B°/B.
Where B° and B are the amount of the initial concentration and the amount of the concentration remaining, k is the rate constant and t = time taken for the concentration to decrease.
So, we have; time taken,t = ln( 0.860/.310)/0.500.
==> ln 2.77/0.500.
==> time taken,t =2.038 seconds.
Cocoa butter, the fat in chocolate, can crystallize in any one of 6 different forms (polymorphs, as they are called). Unfortunately, only one of these, the beta crystal (or Form V), hardens into the firm, shiny chocolate that cooks want. Form VI is also a stable hard crystal, but only small amounts of it form from the good beta (Form V) crystals upon lengthy standing. When you buy commercial chocolate it is in the form of beta crystals.
When you melt chocolate and get it above 94° F, you melt these much desired beta crystals and other types of crystals can set up. If you simply let melted chocolate cool, it will set up in a dull, soft, splotchy, disgusting-looking form. Even the taste is different. Fine chocolate has a snap when you break it and a totally different mouthfeel from the other cocoa butter forms.
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