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kotegsom [21]
3 years ago
10

Does the nucleus (protons and neutrons) weigh more than the electrons?

Chemistry
2 answers:
Butoxors [25]3 years ago
3 0
The mass of nucleons (and thus of the nucleus) is roughly 1000 times greater than that of electrons.
vichka [17]3 years ago
3 0
Yes, protons and neutrons weigh much more than electrons. Which is why the mass of an element is the protons and neutrons added together, disregarding the electrons. Their weight is almost negligible.
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3 years ago
The reaction A + 2B ® products has been found to have the rate law, rate = k[A] [B]2. While holding the concentration of A const
Over [174]

Answer:

The rate of the reaction will increase by a factor of 9.

Explanation:

Hello,

In this case, considering the given second-order reaction, whose rate law results:

r=k[A] [B]^2

We easily infer that at constant concentration of A but tripling the concentration of B, we are going to obtain the following increasing factor while holding the remaining variables constant:

Increase\ factor=\frac{r_{final}}{r_{initial}} =\frac{k[A][3*B]^2}{k[A][B]^2} =\frac{3^2}{1} \\Increase\ factor=9

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3 0
3 years ago
1. Choose the correct answer from the given alternative. 1. A blood sample analysis showed that a total of 0.0000036g lead. Expr
son4ous [18]

Answer:

3.6 micrograms answers (c)

Explanation:

5 0
2 years ago
A solution of h2so4(aq) with a molal concentration of 2.24 m has a density of 1.135 g/ml. what is the molar concentration of thi
OLga [1]
This problem is simply converting the concentration from molality to molarity. Molality has units of mol solute/kg solvent, while molarity has units of mol solute/L solution.

2.24 mol H2SO4/kg H2O * (0.25806 kg H2SO4/mol H2SO4) = 0.578 kg H2SO4/kg H2O

That means the solution weighs a total of 1 kg + 0.578 kg = 1.578 kg. Then, convert it to liters using the density data:

1.578 kg * (1000g / 1kg) * (1 mL/1.135 g) = 1390 mL or 1.39 L.

Hence, the molarity is

2.24/1.39 = 1.61 M
7 0
3 years ago
Please help make sure its correct thanks
Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g

Then mol iodine (MW=126.9045 g/mol) :

\tt \dfrac{84.1}{126.9045}=0.663

mol ratio of Cobalt and Iodine in the compound :

\tt 0.221\div 0.663=1\div 3

5 0
2 years ago
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