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3 years ago

Which mineral might scratch the mineral fluorite but would not scratch the mineral amphibole

Chemistry

2 answers:

morpeh [17]3 years ago

8 0

Gypsum is the mineral

omeli [17]3 years ago

8 0

Answer: Apatite

Explanation: According to the Mohr's hardness scale, fluorite has a hardness of 4, whereas amphibole (which belongs to the hornblende field) has a hardness of about 5-6. And the hardness of apatite is 5.

So by comparing their hardness, we can determine that Apatite can easily scratch the mineral fluorite because it has higher hardness than fluorite, but apatite will not be able to scratch amphibole because apatite has a lower hardness compared to amphibole.

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Most acid precipitation results from the combination of _____ with water in the atmosphere, forming strong acids that fall with

aleksley [76]

Answer:

Most acid precipitation results from the combination of <u>Sulfur Oxides and Nitrogen Oxides</u> with water in the atmosphere, forming strong acids that fall with rain or snow.

Explanation:

Acid rain is said to be that rain which contains high concentration of H⁺ ions. The main source of acid rain is the elimination of NOₓ (Nitrogen Oxides) and SOₓ (Sulfur Oxides) from different means in industries and other combustion processes on earth.

Examples:

SO₂ + H₂O → H₂SO₄

NO₂ + OH° → HNO₃

From above examples it can be seen that the sulfur and nitrogen oxides when reacted with water forms strong acids. These acids come along with rain water and causes different problems to living organisms and non living objects like buildings.

4 0

3 years ago

What is the name of the compound Ca(NO3)2 ?

natulia [17]

Answer: B) calcium nitrate

5 0

3 years ago

When an acid reacts with a strong base which product always forms

patriot [66]

Answer:طيزي

Explanation:

4 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$