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zaharov [31]
3 years ago
10

Resonance in air columns is used in which of these musical Instruments?

Chemistry
2 answers:
AnnZ [28]3 years ago
5 0
Resonance in air columns is used in a variety of musical woodwind instruments. Something like a flute for example uses resonance to make sound. Not sure what instruments you are talking about unless you provide a list.
Reptile [31]3 years ago
3 0

Answer: Trumpets makes the most sense

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1. How many molecules of H,O are in 4.32 moles?
Korvikt [17]

Answer:

dont know

Explanation:

8 0
3 years ago
A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta
melomori [17]
Cao +  H2O  ---->Ca(OH)2
Calculate   the  number  of  each   reactant  and  the  moles  of  the  product
that  is
moles = mass/molar mass
The  moles  of  CaO=  56.08g/  56.08g/mol(molar  mass  of  Cao)=  1mole
the  moles  of  water=  36.04 g/18  g/mol=  2.002moles
The   moles  of Ca (OH)2=74.10g/74.093g/mol= 1mole

 The  mass  of differences  of  reactant  and  product  can   be  therefore 
 explained  as 
 1  mole   of  Cao  reacted  completely   with   1  mole   H2O  to  produce  1 mole  of  Ca(OH)2. The  mass  of  water   was  in  excess  while  that  of  CaO  was  limited

3 0
3 years ago
HELPP ASAPP..........
Ksivusya [100]

Answer:

A, C, D.

Explanation:

6 0
3 years ago
Read 2 more answers
What percentage of a radioactive species would be found as daughter material after six half–lives?
Novosadov [1.4K]
100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................

After six half-lives would be found 1.5625% of readioactive species.
3 0
4 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
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