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nikklg [1K]
4 years ago
9

Sometimes rock formations form because pieces of the rock has been crumbled and been blown/washed away. This is and example of w

hat?
Chemistry
2 answers:
butalik [34]4 years ago
6 0
Hi. I think it's erosion.
liubo4ka [24]4 years ago
5 0
Mechancial weathering i think

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What is a conversion factor?
Zarrin [17]

Answer: an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

Explanation:

5 0
3 years ago
Read 2 more answers
Calculate the number of molecules in 7 moles of oxygen
Alexxandr [17]

Answer:

4.21

Explanation:

use Avogadro's number

6.023 x 10^23

multiply this by 7 because you want to find 7 moles :

6.023 x 10^23 x 7 = 4.21

3 0
3 years ago
The balanced equation below shows the products that are formed when butane (C4H10) is combusted.
Nataly [62]

Answer:

2:8

Explanation:

The reaction equation is a given as:

         2C₄H₁₀   +    130₂   →    8CO₂     +     10H₂O  

From the reaction equation, the mole ratio is 2:8

Butane is C₄H₁₀

Carbon dioxide CO₂

From the reaction;

       2 moles of butane will produce 8 moles of carbon dioxide

3 0
3 years ago
Read 2 more answers
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

The volume of solution is 225 mL, converting this into L,

1 mL=10^{-3}L

Thus,

225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

Therefore, concentration of potassium sulfate solution is 0.236 M.


4 0
3 years ago
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