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2 years ago

12

I need help with science pls help me

1 answer:

ryzh [129]2 years ago

6 0

Answer:

Explanation:

6. When insulating materials rub against each other, they may become electrically charged .

7. Charging by conduction involves the contact of a charged object to a neutral object.

8. Grounding is the process of removing the excess charge on an object by means of the transfer of electrons between it and another object of substantial size.

9. Grounding is the process of removing the excess charge on an object by means of the transfer of electrons between it and another object of substantial size.

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The table compares some characteristics of two atoms.

Mademuasel [1]

Answer:

Explanation:

Atom X,

Neutron num = 4, Mass num = 7

Therefore, proton number = Mass num - Neutron num

= 7 - 4

= 3

Atom Y,

Neutron num = 5, Mass num = 9

Therefore, proton number = Mass num - Neutron num

= 9 - 5

= 4

1) False. Atom X belongs to group 3 while atom Y belongs to group 4

2) False. Atom X is in a column on the left of Atom Y

3) True. Atom X and atom Y belong to the same period two on the periodic table

4) Fasle. Atom X and Y are not isotopes because they do not have the same atomic number.

Isotopy is a phenomenon whereby atom of same element have same atomic number but different mass number

4 0

3 years ago

The first question is the one I need help with

aleksandr82 [10.1K]

I looked at the question and there isn't a lot of information. im assuming the answer is carbon dioxide. its the only substance that would make sense I hope it helps.

5 0

3 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

2 years ago

Which statement is true?

dimulka [17.4K]

Answer:

A I think

Explanation:

6 0

3 years ago

A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated nitric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted nitric acid solution

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

Hence, the volume of concentrated solution required is 9.95 mL

6 0

2 years ago