What are the percent compositions of C and H in C2H2?

1. Is air a mixture or a compound?<br> mixture

Mekhanik [1.2K]

Its mixture goodluck :)

8 0

2 years ago

Which is stronger - the attractive forces between water molecules and chromium and chloride ions, or the combined ionic bond str

Allisa [31]

Out of the two, the forces between water molecules and chromium and chloride ions is greater. This is proven by the fact that chromium chloride is slightly soluble in water, about 565 grams per liter.
In order for a substance to be soluble, the attraction of the ions to the water molecules must exceed the attraction between its own molecules and the water molecules.

3 0

3 years ago

A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib

DochEvi [55]

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

$I_2+Br_2\rightarrow 2IBr$

It is possible to set up the following equilibrium expression:

$K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110$

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of $x$ (reaction extent) would be:

$0.0110=\frac{(2x)^2}{(0.50-x)^2}$

Which can be solved for $x$ to obtain two possible results:

$x_1=-0.0277M\\\\x_2=0.0245M$

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

$[IBr]=2x=2*0.0249M=0.049M$

Regards!

7 0

2 years ago

Question 3 (1 point)

Hoochie [10]

Answer:

50 mol

Explanation:

Mass of methane = 800 g

Number of moles of CO₂ produced = ?

Solution:

Chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Number of moles of methane:

Number of moles = mass/molar mass

Number of moles = 800 g/ 16 g/mol

Number of moles = 50 mol

Now we will compare the moles of methane and carbon dioxide from balanced chemical equation.

CH₄ : CO₂

1 : 1

50 : 50

4 0

3 years ago

Carry out the following operations as if they were calculations of experimental results and express each answer in standard nota

Shalnov [3]

Answer:

1. $10.6\; \rm m$ (one decimal place.)

2. $0.79\; \rm g$ (two decimal places.)

3. $16.5\;\rm cm^2$ (three significant figures.)

Explanation:

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

$5.6792\;\rm m$ has four decimal places.
$0.6\; \rm m$ has only one decimal place.
$4.33\; \rm m$ has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: $\rm m$ .)

Therefore:

$\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m$ . (Rounded to one decimal place.)

Similarly:

$\rm 3.70\; \rm g$ has two decimal places.
$2.9133\; \rm g$ has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be $\rm g$ (same as the unit of the two inputs.)

$\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g$ . (Rounded to two decimal places.)

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

$4.51\; \rm cm$ has three significant figures.
$3.6666\; \rm cm$ has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be $\rm cm \cdot cm$ , which is occasionally written as $\rm cm^2$ .

$\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2$ . (Rounded to three significant figures.)

7 0

3 years ago