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aivan3 [116]
4 years ago
10

The phenomenon of marble cancer is due to (a) soot particles (c) fog (b) CFCs (d) acid rain

Chemistry
1 answer:
Scorpion4ik [409]4 years ago
3 0

<u>Answer</u>:

The phenomenon of marble cancer is due to ACID RAIN.

<u>Explanation</u>:

The marble corrosion is a process by which the marbles or limestone used in construction of monuments and buildings is corroded away by the acid rain.  The sulphur-dioxide and  nitrogen oxides present in the atmosphere undergoes certain chemical reaction causing this acid rain. These chemical in the acid rain converts the calcium  present in the limestone or marble into salt.

Also, this marble cancer leads to change the appearance and color. Sometime due to over corrosion the monuments or building causes them disintegrate .Tajmahal is one of the monument affected by marble cancer. the white color of tai mahal is slowly turning into yellow due to this marble cancer.

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Consider the reaction 3X + 2Y → 5C + 4D
Aloiza [94]

This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the moles of X that are needed to produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:

<h3>Mole ratios:</h3><h3 />

In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.

<h3>Calculations:</h3>

In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:

21.00molD*\frac{3molX}{4molD} =15.75molX

Learn more about mole ratios: brainly.com/question/15288923

7 0
3 years ago
How much aluminum oxide in grams is produced from the reaction of 3.5 moles of oxygen with 4.5 moles of aluminum?
bagirrra123 [75]

Answer:

The mass of 3.5 moles of Ca is 140 g to two significant figures

7 0
3 years ago
A cooling of surface ocean waters in the eastern equatorial Pacific Ocean is known specifically as ________.
Eddi Din [679]

Answer:

La Nina

Explanation:

La Nina -

The word La Nina means " the little girl " , in spanish language .

La Nina , is an ocean - atmospheric process , which occurs in the eastern equatorial Pacific Ocean .

It is exactly opposite of El Nino , which means " the little boy " .

In the phase of La Nina , the temperature of the eastern equatorial Pacific Ocean becomes lower than the usual temperature by around 3°C to 5°C .

This phase of La Nina is for around 5 months in an year .

Hence , from the given statement of the question ,

The correct term is La Nina .

8 0
4 years ago
For the following reaction, 42.2 grams of potassium hydrogen sulfate are allowed to react with 21.4 grams of potassium hydroxide
ASHA 777 [7]

Answer:

53.99g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

KHSO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Step 2:

Determination of the masses of KHSO4 and KOH that reacted and the mass of K2SO4 produced from the balanced equation.

This is illustrated below:

Molar mass of KHSO4 = 39 + 1 + 32 + (16x4) = 136g/mol

Mass of KHSO4 from the balanced equation = 1 x 136 = 136g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Mass of KOH from the balanced equation = 1 x 56 = 56g

Molar mass of K2SO4 = (39x2) + 32 + (16x4) = 174g/mol

Mass of K2SO4 from the balanced equation = 1 x 174 = 174g.

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH to produce 174g of K2SO4

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH.

Therefore, 42.2g of KHSO4 will react with = (42.2 x 56)/136 = 17.38g of KOH.

From the above calculations, we can see that only 17.38g out of 21.4g of KOH given was needed to react completely with 42.2g of KHSO4.

Therefore, KHSO4 is the limiting reactant and KOH is the excess reactant.

Step 4:

Determination of the maximum mass of K2SO4 produced from the reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction. The limiting reactant is KHSO4 and the maximum amount of K2SO4 produced can be obtained as follow:

From the balanced equation above, 136g of KHSO4 reacted to produce 174g of K2SO4.

Therefore, 42.2g of KHSO4 will react to produce = (42.2 x 174)/136 = 53.99g of K2SO4.

Therefore, the maximum amount of K2SO4 produced is 53.99g.

8 0
3 years ago
According to this Bohr model, in what period would this element be found?
Tasya [4]

Answer:

yes

Explanation:

yes

4 0
3 years ago
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