All categories

2 years ago

BRAINLIESIT IF CORRECT 8

Chemistry

2 answers:

Oksi-84 [34.3K]2 years ago

7 0

Answer:

I think ur answer would be refracted.

Explanation:

Paladinen [302]2 years ago

6 0

Answer:

refracted

Explanation:

You might be interested in

Use the word bank to complete the charts about the metrics system measurement

Aloiza [94]

Answer:

Length

- centimetre

- millimetre

- kilometre

- meter

Mass

- gram

- kilogram

Capacity

- liter

- millilitre

I don't know why mass is there but hope this helps! :)

5 0

3 years ago

What is a catalyst?

luda_lava [24]

a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.

- a person or thing that precipitates an event.

7 0

3 years ago

Read 2 more answers

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

3 years ago

What do elements in the first two columns of the periodic table have in common?

coldgirl [10]

Answer: b.) they tend to lose electrons to gain stability

Explanation:

5 0

2 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

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

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

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

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

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

2 years ago