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3 years ago

Soil shows what evidence of life

1 answer:

5 0

oils are complex mixtures of minerals, water, air, organic matter, and countless organisms that are the decaying remains of once-living things. It forms at the surface of land – it is the “skin of the earth.” Soil is capable of sup- porting plant life and is vital to life on earth.

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Calculate the molarity of 0.700 mol of na2s in 1.05 l of solution.

slamgirl [31]

Molarity is the amount of solute molecule (in moles) per 1L of solvent. In this case, the solute is 0.7mol Na2S and the solvent volume is 1.05L. Since the unit in this problem is already mol and L then you don't need to do any conversion of the units. The calculation would be:

molarity = mol of solute / (1L/ volume of solvent)
molarity = 0.7 mol/ (1L/ 1.05L)= 0.67M

6 0

2 years ago

Consider the balanced equation for the following reaction:

Zanzabum

<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 5.663 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0

2 years ago

Use the following information to determine the empirical formula if a compound is found to have 18.7% Li, 16.3% C, and 65.5% O A

Elina [12.6K]

Answer:

The empirical formula is Li2CO3

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100 grams

The compound contains :

18.7% Li = 18.7 grams of Li

16.3 % C = 16.3 grams of C

65.0% O = 65.0 grams of O

Total = 100%

Molar mass of Li = 6.94 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles

Moles = Mass / molar mass

Moles Li = 18.7 grams / 6.94 g/mol = 2.65 moles

Moles C = 16.3 grams / 12.01 = 1.36 moles

Moles O = 65.0 grams / 16.0 g/mol = 4.06 moles

Step 3: Calculate the mol ratio

We divide by the smallest number of moles

Li: 2.65/1.36 ≈ 2

C: 1.36/1.36 = 1

O: 4.06/1.36 ≈ 3

The empirical formula is Li2CO3

3 0

3 years ago

Convert .0027 CL to mL.

Over [174]

Answer:

0.027 ml

Explanation:

$1\:cl = 10\:ml\\0.0027 \: cl = x\:ml$

Cross Multiply

$x = 0.027 \:ml$

4 0

2 years ago

What type(s) of orbital overlap is(are) used to form the indicated bond in the following structure.

lorasvet [3.4K]

D.....................

3 0

2 years ago