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3 years ago

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How do the molecules differ under the microscope when it becomes winter?What state is the water in at this time of year? Please

help me thank you

1 answer:

EleoNora [17]3 years ago

8 0

Answer:

They come together and freeze. They turn into a solid and turn into ice.

Explanation:

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Calculate the molecular (formula) mass of each compound: (a) iron(ll) acetate tetrahydrate; (b) sulfur tetrachloride; (c) potass

Nutka1998 [239]

Answer:

a) Iron(ll) acetate tetrahydrate: 245,68 g/mol

b) Sulfur tetrachloride: 173,87 g/mol

c) Potassium ermanganate 158,034 g/mol

Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

The molecule of Iron(II) acetate itetrahydrate is (CH3COO)2Fe•4H2O, it means that you have:

2 atoms of carbon times the atomic weight of C (12.00g/mol)= 24g

14 atoms of Hidrogen times the atomic weight of H (1,00g/mol)= 14g

6 atoms of Oxigen times the atomic weight of O (16,0g/mol)= 96

2 atoms of Iron times the atomic weight of Fe (55,84g/mol)= 111,68g

At last, you only have to add the results: 24+14+96+11,68= 245,68g/mol. This example was for the first molecule.

See you,

5 0

3 years ago

in rutherford's gold foil experiment a very small number of alpha particles were deflected. what about the structure of the atom

zvonat [6]

Answer:The atom being mostly empty space. A small number of alpha particles were deflected by large angles (> 4°) as they passed through the foil. There is a concentration of positive charge in the atom. Like charges repel, so the positive alpha particles were being repelled by positive charge

8 0

3 years ago

The idea that light can act like packets led to what new field on science?

Elenna [48]

Light acting as 'packets' of exact amounts of energy (a particle-like quality) called quanta led to the development of quantum mechanics. Light also has wave qualities (wavelength, frequency, amplitude) which is referred to as particle-wave duality.

5 0

3 years ago

Read 2 more answers

A stream of surface water reaches a porous portion of sediment and seeps into the ground. This water eventually joins a large re

Fed [463]

The correct answer is - C. Hydrosphere; geosphere.

The hydrosphere is the sphere which contains all the waters on the planet Earth. The geosphere is the sphere that contains the solid rocky part of the Earth. When the water goes through the porous sediments and enters deeper into the ground, that means the we have an interaction of water and rocks. In other words, since the water is part of the hydrosphere, and the rocks are part of the geosphere, we have an interaction between the hydrosphere and the geosphere.

8 0

4 years ago

Read 2 more answers

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi

avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction : $Ni^{2+}+2e^-\rightarrow Ni$

The expression for reaction quotient will be :

$Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(0.0141)}{(0.00104)}=13.6$

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

$E^o_{cell}$ = standard electrode potential of the cell = 0.51 V

$E_{cell}$ = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

$E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)$

$E_{cell}=0.47V$

Therefore, the actual cell potential of the cell is 0.47 V

4 0

4 years ago