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Bas_tet [7]
3 years ago
12

How many moles are in 12 grams of potassium chlorate, KClO3?

Chemistry
1 answer:
galina1969 [7]3 years ago
6 0
It has one mole xxxxxxxx
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How many liters of 1.2M solution can be prepared with 0.50 moles of C6H12O6
BigorU [14]
First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
5 0
3 years ago
For many purposes we can treat ammonia (NH) as an ideal gas at temperatures above its boiling point of -33.C. Suppose the temper
levacccp [35]

Answer:

In homeothermic (“warm-blooded”) animals, body temperature is carefully  

regulated. The hypothalamus, located in the brain, acts as the master ther-

mostat to keep body temperature constant to within a fraction of a degree  

Celsius in a healthy animal. If the body temperature starts to deviate much  

from the desired constant level, the hypothalamus causes changes in blood  

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the temperature back to normal. What evolutionary advantage does a con-

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over the poikilotherms (e.g., reptiles and insects), whose body temperatures  

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Explanation: Basic chemical understanding as revealed upwards  

6 0
3 years ago
Calculate the mass of 3 moles of clorine​
Vitek1552 [10]
Moles Chlorine to grams = 35.453 grams.
moles Chlorine to grams = 70.906 grams.
moles Chlorine to grams = 106.359 grams.
moles Chlorine to grams = 141.812 grams.
moles Chlorine to grams = 177.265 grams.
moles Chlorine to grams = 212.718 grams.
4 0
2 years ago
Write a balanced nuclear equation for the formation of 28 Si 14 from beta-minus emission.
Fantom [35]

Answer:

Phosphorus-28 undergoes beta-minus decay to produce

  • an electron,
  • a Silicon-28 nuclei, and
  • an electron antineutrino.

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}

Explanation:

In simple words, when a nucleus undergoes beta-minus decay, a neutron is converted to a proton. An electron and an electron antineutrino will be released.

\rm ^{1}_{0}n^{0} \to ^{1}_{1}p^{+} + ^{\phantom{-}0}_{-1}e^{-} + \bar{\mathnormal{v}}_{\rm e}.

One way to tell whether a neutron is converted to a proton, but not vice versa, is to check the sum charges on the two sides of this equation.

  • Left-hand side: 0. Neutron is neutral.
  • Right-hand side: 1 + (-1) = 0. Each proton carries a charge of +1. Each electron (beta-minus particle) carries a charge of -1. Antineutrinos are neutral.

The charges on the two sides of this equation is the same. Hence this nuclear equation is possible (but not necessarily correct; however, if the proton and the neutron are in the wrong place the charge won't even be the same.)

Since the mass number of a proton and a neutron are both 1, the overall mass number of the atom will stay the same.

The atomic number is the number of protons in each atom. That number determines the symbol and the chemical properties of the atom. When one neutron in an atom is converted to a proton, the atomic number of the atom will increase by 1.

The atomic number of the daughter nucleus, silicon, is 14. It takes a parent nucleus with atomic number 14 - 1 = 13 to produce a silicon atom. Refer to a modern periodic table. Atomic number 13 corresponds to the element aluminum.

Also, the mass number of the daughter nucleus is 28. Since the mass number would stay the same in a beta decay, the mass number of the parent nucleus would also be 28. In other words, it takes an aluminum-28 atom to undergo beta-decay to produce a silicon-28 atom.

Complete the other details (electron and electron antineutrino) to obtain the equation

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}.

6 0
3 years ago
How can I find the answer for this ?
12345 [234]
For what? what is your question 
7 0
3 years ago
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