-Before we begin each flame test we will

The chemical formula for table sugar is C12H11O22. What can you tell from this formula?

just olya [345]

I say the answer is The ratio of oxygen atoms to hydrogen atoms in a molecule of sugar is 2 to 1

7 0

3 years ago

These goats are eating grass on a sunny day. What is happening to the carbon in the air around the living things on the mountain

polet [3.4K]

Answer:

both

Explanation:

Carbon in the air around the living thing is moving in and out of its lungs. The movement is occurring at the same time. This is one of the most important gaseous exchange important to life.

The goats takes in oxygen gas from the surrounding and releases carbon dioxide in the process.
But, air is actually drawn in by the goat which is a mixture of several gases.
Air contains carbon dioxide which is a rich source of carbon
With the carbon dioxide from respiratory processes, the goat ejects and breathes out this waste carbon matter.
Therefore, the gaseous exchange in a goat involves the movement of carbon in and out of the air.

8 0

3 years ago

Calculate the mass of 3.5 mol C6H6

Lilit [14]

(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6

4 0

3 years ago

Which of the following are evidence that chemical change has occurred

dolphi86 [110]

Answer:

gas given off

Explanation:

that's the answer

3 0

3 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago