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IgorLugansk [536]
2 years ago
7

During an ionic bond, electrons are basically transferred from one atom to another whereas in a covalent bond, electron pairs ar

e shared. true or false
Chemistry
1 answer:
Ad libitum [116K]2 years ago
8 0
The answer is true......

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At their centers, all the jovian planets have cores made of:
Paul [167]

Answer:

The core of Jupiter and Saturn is made up of rock,metal and hydrogen compounds, while the core of Uranus and Neptune is made up of rock,metals,water, methane, and ammonia.

Explanation:

The Jovian planets include, Jupiter, Saturn, Uranus, and Neptune. These planets when compared to terrestrial planets are small, with dense cores and surrounded by layers of gas.

5 0
2 years ago
Equal amounts of N2 and O2 are added, under certain conditions, to a closed container. Which changes occur in the reverse reacti
geniusboy [140]
1) The forward reaction is N2 (g) + O2 (g)  → 2NO

(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.

2) The equiblibrium equation is

  N2 (g) + O2 (g)  ⇄ 2NO

3) Then, the reverse reaction is

2NO → N2(g) + O2(g)

Answer: 2NO → N2(g) + O2(g)
4 0
3 years ago
What is the scientific method?
AysviL [449]
C. a plane for forming a scientific hypotheses
3 0
3 years ago
There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic
Leya [2.2K]
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The statements that explains this phenomenon are:
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6 0
2 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
3 years ago
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