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Paha777 [63]
3 years ago
10

The image above shows the nucleus of a nitrogen atom. How can the atomic number of nitrogen be determined?

Chemistry
2 answers:
Anettt [7]3 years ago
7 0

Answer:

A

Explanation:

Julli [10]3 years ago
3 0
I think you forgot to attach the diagram along with the question. I am answering the question based on my research and knowledge. "Count the number of protons" is the one among the following choices given in the question by which the <span>atomic number of nitrogen be determined. The correct option among all the options that are given in the question is the second option or option "B".</span>
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Write the balanced equation for the neutralization reaction between HCl and Ba(OH)2 in an aqueous solution. Phases are optional.
ad-work [718]

Explanation:

2HCl + Ba(OH)2 ------> BaCl2 + 2H2O

4 0
2 years ago
Which of the following representations shows the correct placement of trophic levels?
OleMash [197]
You have to start listing from the bottom :

3. Secondary Consumers

2. Primary Consumers

1. Producer

4 0
3 years ago
Find the mass, in grams, of 3.758 mol CH4. ( show work )
vladimir2022 [97]

Answer:

The mass of CH4 is 60, 29 grams.

Explanation:

We use the weight of the atoms C and H for calculate the molar mass:

Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=

Weight of CH4 =16, 042 g/mol

1molCH4-----16, 042grams

3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams

5 0
3 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
What volume of solution gives the desired moles?
motikmotik

Answer:

1. 0.073L

2. 0.028L

3. 0.014L

Explanation:

The volume for the different solutions are obtained as shown below:

1. Mole = 0.53mol

Molarity = 7.25M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.53/7.25

Volume = 0.073L

2. 0.035mol from a 1.25M

Mole = 0. 035mol

Molarity = 1.25M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.035/1.25

Volume = 0.028L

3. Mole = 0.0013mol

Molarity = 0.090M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.0013/0.090

Volume = 0.014L

8 0
3 years ago
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