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Vlada [557]
3 years ago
5

What is relief and how does it affect soil formation

Chemistry
1 answer:
liubo4ka [24]3 years ago
7 0
<span><span>S is for soil,</span><span>cl (sometimes c) represents climate,</span><span>o organisms including humans,</span><span>r relief,</span><span>p parent material, or lithology, and</span><span>t time.</span></span>
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In the process of making soap, I poured some of the cooked mixture through some muslin fabric, in order to separate the solid pa
lyudmila [28]

Answer:

filtering

Explanation:

you're pouring the mixture through muslin cloth to keep the particles and bigger peaces out of the soap.

6 0
3 years ago
Why does the chemical reaction seen here obey the law of conservation of matter?
ss7ja [257]

Why does the chemical reaction seen here obey the law of conservation of matter?

<u><em>Answer:</em></u>

  • Because there are the same number of atoms of each element shown on both sides

<u><em>Explanation</em></u>

  • As in chemical reactions, atoms bonds are break and new bonds are formed. As new substance are formed but overall they have same elements, no new elements come from outside or go to outside. In other words , rearrangement of atoms take place but number of atoms remained same.

                            NaOH + HCl -----> NaCl + H2O

  • As in above reaction there are the same number of atoms of each element shown on both sides .
7 0
3 years ago
Read 2 more answers
Household items that are substances
Yuri [45]

Answer:

water,salt are the house hold items

4 0
4 years ago
What's mole in 34g of H2O<br><br><br>​
enyata [817]

Answer:

I hope you have the answer now

Explanation:

u an

3 0
3 years ago
In a hydrogen atom, the electron and the proton are seperated by about 0.5 angstroms, 5.0×10−11 m. what is the acceleration of a
nadya68 [22]

The charge of proton and electron is equal in magnitude that is 1.6\times 10^{-19} C. The distance between electron and proton is 0.5 \AA or 5\times 10^{-11} m.

Force exerted on electron by proton is calculated as follows:

F= \left | \frac{kq_{e}q_{p}}{r^{2}} \right |

Here, k is Coulomb's constant, q_{p} is charge on proton,  q_{e} charge on electron and r is distance between them.

Putting the values,

F= \left | \frac{(8.99\times 10^{9}N m^{2}/C^{2})(1.6\times 10^{-19} C)(1.6\times 10^{-19} C)}{(5\times 10^{-11} m)^{2}} \right |=9.2\times 10^{-8} N

Now, according to second law of Newton,

F=m_{e}a

Here, m_{e} is mass of electron and a is acceleration.

Also, 1 N=1 kgm/s^{2}

Rearranging,

a=\frac{F}{m_{e}}=\frac{9.2\times 10^{-8} Kg m/s^{2}}{9.11\times 10^{-31} kg}=1.01\times 10^{23} m/s^{2}

Thus, the acceleration of an electron in this atom is 1.01\times 10^{23} m/s^{2}.


7 0
4 years ago
Read 2 more answers
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