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2 years ago

I might honestly fail miserably on my chemistry final tomorrow. Would anyone be able to give some last-minute advice?

Chemistry

1 answer:

Art [367]2 years ago

4 0

Just do the best you can. That’s all that matters in the end!

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Find the mass in kilograms of the liquid air that is required to produce 600L of oxygen. In normal condition, 1L of liquid air h

IgorC [24]

Mass in kilograms of liquid air required = 0.78 kg

Given that

1 Litre of liquid air contains 1.3 grams of oxygen ( air )

Determine the amount of liquid air in Kg

volume of air given = 600 L

mass of liquid air required = x

1 litre = 1.3 grams

600 L = x

∴ x ( mass of liquid air ) = 1.3 * 600

= 780 g = 0.78 kg

Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg

Learn more about liquid air : brainly.com/question/636295

3 0

2 years ago

Read 2 more answers

Molar mass c3h8

maksim [4K]

Answer:

1. 44.11 g

2. 36.03 g

3. 8.08 g

4. 81.7%

5. 18.3%

Explanation:

1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11

2. 12.01×3= 36.03

3. 1.01×8= 8.08

4.(36.03/44.11)×100= 81.7%

5. (8.08/44.11)×100= 18.3%

8 0

2 years ago

Help me get this right no links

ira [324]

Answer:

the rise and fall is the tides.

6 0

3 years ago

Both hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you open

wlad13 [49]

Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass.

Explanation:

4 0

3 years ago

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

Butoxors [25]

Answer:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

$BiO_3^-\rightarrow Bi^{3+}$

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

$(Bi^{5+}O_3)^-\rightarrow Bi^{3+}$

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

$6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O$

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

$6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Then, we accommodate the waters to obtain:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Best regards!

7 0

3 years ago