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2 years ago

Strip mining is less harmfull to the environment than shaft mining but more dangerious to minders

Chemistry

2 answers:

lukranit [14]2 years ago

7 0

I'd say shaft mining I'm 90% sure of it. Hope this helps

yulyashka [42]2 years ago

5 0

False is the correct answer

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Convert 2.17 x 10 -8 Mg into pg

viktelen [127]

This answer is 24 because 2.17 x 10 -8 is 24 so that would be your answer

4 0

3 years ago

What is the molarity of 2.00 L of a solution that contains 14.6 g NaCl?

White raven [17]

Answer: The molarity of the solution is 0.125 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

moles of $NaCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{14.6g}{58.5g/mol}=0.250mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.250mol}{2.00L}=0.125M$

Therefore, the molarity of the solution is 0.125 M

8 0

3 years ago

What adaptations allow a camel and a cactus to survive in warm environments?

lesya [120]

Answer:

A camel stores fat in its hump, while the cactus stores water in its thick stem.

Explanation:

8 0

2 years ago

Read 2 more answers

The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO

mixas84 [53]

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = $\frac{7.5 g}{74.69 g/mol}=0.10 mol$

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

$n=M\time V=6mol/L\times 0.050 L =0.3 mol$

$NiO + H_2SO_4\rightarrow NiSO_4 + H_2O$

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of $NiSO_4$ .

Then 0.10 moles of NiO wil give :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of $NiSO_4$ .

Molar mass of $NiSO_4$ = 154.75 g/mol

Mass of 0.10 moles of $NiSO_4$ :

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of $NiSO_4$ = 15.475 g

Experimental yield of $NiSO_4$ = 17.4 g

Percentage yield :

$\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100$

Percentage yield of $NiSO_4$ :

$\Yield=\frac{17.4}{15.475 g}\times 100=112\%$

Moles of $NiSO_4.6H_2O$ = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of $NiSO_4.6H_2O$ = 17.4 g

Percentage yield of $NiSO_4.6H_2O$ :

$\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%$

3 0

2 years ago

In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par

Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0

3 years ago