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3 years ago

Which subatomic particles contribute to an atom's mass number but not its atomic number

Chemistry

1 answer:

Anarel [89]3 years ago

4 0

The subatomic particle that contributes to an atom's mass number but not to its atomic number would be the neutron.

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Compound 2 contains 2.0g of hydrogen and 32.0g oxygen. What is the percent compound of each element?

Dmitriy789 [7]

Note down the formula below

$\boxed{\sf Mass\%\;of\; element=\dfrac{Mass\:of\:the\: element}{Mass\;of\:the\: compound}\times 100}$

Mass of the compound

$\\ \sf\longmapsto 32+2=34g$

Mass % of Hydrogen:-

$\\ \sf\longmapsto \dfrac{2}{34}\times 100$

$\\ \sf\longmapsto \dfrac{1}{17}\times 100$

$\\ \sf\longmapsto 5.8\%$

Mass % of Oxygen:-

$\\ \sf\longmapsto \dfrac{32}{34}\times 100$

$\\ \sf\longmapsto \dfrac{16}{17}\times 100$

$\\ \sf\longmapsto 94.2\%$

7 0

3 years ago

The element tin has the following number of electrons per shell: 2, 8, 18, 18, 4. Notice that the number of electrons in the out

bazaltina [42]

The question is incomplete, the complete question is:

The element tin has the following number of electrons per shell: 2.8. 18, 18, 4. Notice that the number of electrons in the outer shell of a tin atom is the same as that for a carbon atom. Therefore, what must be true of tin? Tin is a polar atom and can bind to other polar atoms. Tin has a high molecular weight to give tin-containing molecules greater stabilty. All of the above Tin conform single covalent bonds with other elements, but not double or triple covalent bonds Tincan bind to up to four elements at a time

Answer:

Tin can bind to up to four elements at a time

Explanation:

Certain important points were made in the question about tin and one of them is that tin is an element in the same group as carbon hence it has the same number of valence electrons as carbon.

Carbon is always tetra valent. The tetra valency of carbon is the idea that carbon forms four bonds.

If tin has the same number of valence electrons as carbon, then, tin can bind to up to four elements at a time

3 0

3 years ago

Predict the products of below reaction, and whether the solution at equilibrium will be acidic, basic, or neutral.

kati45 [8]

Answer: The product of the given reaction is $HNO_{3}$ and the solution at equilibrium will be acidic.

Explanation:

When two or more chemical substances react together then it forms new substances and these new substances are called products.

For example, $3N_{2}O_{5} + 3H_{2}O \rightarrow 6HNO_{3}$

This shows that nitric acid $(HNO_{3})$ is the product formed and it is an acidic substance.

Hence, the solution at equilibrium will be acidic in nature.

Thus, we can conclude that the product of the given reaction is $HNO_{3}$ and the solution at equilibrium will be acidic.

8 0

3 years ago

Need help ASAP please show your work

VikaD [51]

Answer:

q = 14049 J

Explanation:

q = m*c*(t2-t1)

q = 350 * 0.892 * (70-25) =

312.2 * 45 = 14049 J

I might be getting a little confused but I could be right.

Hope this helps!

4 0

3 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago