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Alexxandr [17]
3 years ago
10

If you can help me answer any of these I’d appreciate it so much!

Chemistry
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

I  can answer some of these.

Explanation:

Q 36.  B. Is false.

Q. 37  C. is false.

Q. 28.

All the  numbers are 1.

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Which type of equipment would be used to precisely measure 26.0 mL of dilute hydrochloric acid?
VikaD [51]
The type of equipment that would be used to precisely measure 26.0 mL of dilute hydrochloric acid would be C. 50 mL graduated cylinder.
D doesn't have enough mLs to measure this, and A and B have too many. 
8 0
3 years ago
Read 2 more answers
Please Help me
dangina [55]
Democritus was the first to propose the idea of the atom. He said the atom was just this tiny, solid sphere. However, he used no scientific evidence to support his claim, so a guy named John Dalton did some experimenting and basically backed up Democritus' claim with evidence. Then, a guy named J.J. Thompson came along and said the atom was not solid and that is consisted of tiny negatively charged particles(electrons) and he came up with the Plum Pudding model which is just a tiny sphere with a punch of random scattered dots in it. After that, Ernest Rutherford did experiments and found that the tiny sphere is made up of mostly empty space with a tiny, dense, positively charged sphere inside of it, and the negatively charged particles just randomly float around it. Neils Bohr then said that the electrons take specific, circular, evenly spaced paths. Then, finally, we come to the Quantum Mechanical Model which is the one accepted today. This model basically vetos Bohr's idea and has a nucleus inside of an electron cloud, which is where the electrons are found.
6 0
3 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
If a piece of matter changes shape, how does its mass change?
kap26 [50]

The law of conservation of mass states that matter cannot be created nor destroyed the answer would be D. It remains the same
5 0
3 years ago
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