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3 years ago

11

What is the nucleus of an atom?

1 answer:

Leno4ka [110]3 years ago

5 0

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

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The H30+ concentration of a solution is equal to 0.8 M. What is the pH of the solution?

timurjin [86]

Answer:

option A= pH = 0.1

Explanation:

Given data:

Concentration of H₃O⁺ = 0.8 M

pH = ?

Solution:

pH = -log [H₃O⁺] H₃O⁺ = H⁺

pH = -log [0.8]

pH = - (-0.097)

pH = + 0.1

5 0

3 years ago

Scientific Method

Leona [35]

Answer:

Problem, Variables, Hypothesis, Research, Procedure, Data, Conclusion.

Explanation:

First, you have to know what the problem that you're talking about is. Then, you have to know the independent, dependent, and controlled variables are in order to make a hypothesis. Then do research about the topic. After that, you make the procedure your experiment, if that's what your doing. Then after the experiment, you find the data. Using that data, you make a conclusion.

4 0

3 years ago

HELP ITS DUE IN 5 MINS PLSSS Name at least one negative affect that is the result of clearing areas of a rainforest.

Dimas [21]

Global warming

Carbon dioxide is released into the air, causing global warming.

6 0

3 years ago

Read 2 more answers

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0

3 years ago

How are molecules built?

RoseWind [281]

Bonding I'm guessing

6 0

3 years ago