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2 years ago

I really need help with this question. This is for chemistry

Chemistry

1 answer:

chubhunter [2.5K]2 years ago

8 0

Answer:

B.) 33.5 mL

Explanation:

The density measurement can be used as a conversion. It allows you to go between grams and mL. When you say the density of mercury is 13.6 g/mL, you are technically saying that there are 13.6 grams of mercury per every 1 mL. So, if you want to convert between grams and mL, you need to multiply the staring amount by this conversion. Because you want to cancel out grams, you need to put this unit in the denominator of the ratio.

455 grams mercury 1 mL
------------------------------ x -------------------- = 33.5 mL mercury
13.6 grams

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A sample of CO2 gas has a volume of 2.7 L at 78.5 kPa. At what pressure would this sample of gas have a volume of 4.0L? Temperat

lisabon 2012 [21]

Answer:

52.99 kPa

Explanation:

Initial volume V1 = 2.7 L

Initial Pressure P1 = 78.5 kPa

Final Volume V2 = 4.0L

Final Pressure P2 = ?

Temperature is constant

The relationship between these quantities is given by the mathematical expression of Boyles law. This is given as;

V1P1 = V2P2

P2 = V1P1 / V2

P2 = 2.7 * 78.5 / 4.0

P2 = 52.99 kPa

7 0

3 years ago

Limestone stalactites and stalagmites are formed in caves by the following reaction: Ca2 (aq) 2HCO−3(aq)→CaCO3(s) CO2(g) H2O(l)

Vesna [10]

Answer : The value of $\Delta E$ for this reaction is 36.18 kJ

Explanation :

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

$\Delta E=q+w$

where,

$\Delta E$ = internal energy of the system

q = heat added or rejected by the system

w = work done

As we are given that:

q = 38.65 kJ

w = -2.47 kJ (system work done on surrounding)

Now put all the given values in the above expression, we get:

$\Delta E=38.65kJ+(-2.47kJ)$

$\Delta E=36.18kJ$

Therefore, the value of $\Delta E$ for this reaction is 36.18 kJ

6 0

2 years ago

Compare and contrast the burning of wood and the metabolism of glucose in your cells. How are they similar, and how are they dif

Alisiya [41]

Burning of wood is a combustion reaction and the metabolism of glucose in your cells is cellular respiratory reaction.

Cellular respiration releases stored energy in glucose molecules and converts it into a form of energy that can be used by cells.

Cellular respiration is the process by which organisms combine oxygen with foodstuff molecules, diverting the chemical energy in these substances into life-sustaining activities and discarding, as waste products, carbon dioxide and water.

Wood as well as many common items that combust are organic (i.e., they are made up of carbon, hydrogen and oxygen). When organic molecules combust the reaction products are carbon dioxide and water (as well as heat)

Similarities:

1. Combustion reaction and metabolism of glucose both require oxygen.

2. Combustion and metabolism of glucose both product carbon dioxide and water

3. Both produces by-products: After cellular respiration and combustion have gotten what they needed from the wood, there will be byproducts from the conversion. In the case of combustion, they are noxious gases like carbon monoxide. In the case of respiration, the sugar molecule is broken into two molecules of pyruvic acid.

4. Catalyst: While breaking apart the bonds to release the stored energy either combustion or sugars for respiration the bonds will not broken by themselves. In each case, a catalyst is required to start the reaction that will break the bonds apart. In the case of combustion, the catalyst is a spark. Wood are flammable, so the spark will ignite the burning, breaking apart the bonds and releasing the energy. For respiration, enzymes are used to break the sugar molecule apart.

Differences

1. Glucose metabolism produces a chemical energy, while combustion produces light and heat energy.

2. Glucose metabolism is an endothermic reaction while combustion is an exothermic reaction (produces heat)

8 0

3 years ago

Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the

Mekhanik [1.2K]

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12". Where 18 is the number of carbon atoms, 2 is the number of double bonds, and the numbers successive to Δ "delta" the position of the double bonds starting to count from the carboxylic -COOH end of the molecule.

In other words, the main functional group is a carboxylic acid. We have a total of 18 carbons. Additionally, we have 2 double bonds. On carbons 9 and 12.

Lets see figure 1

I hope it helps!

5 0

3 years ago

In three to five sentences, identify two components of the control subsystem of a vehicle, and use them

tekilochka [14]

Answer:

The control subsystem of a vehicle can have several parts. Two examples are the braking and steering wheel. If the subsystem were to malfunction, driving could potentially be dangerous due to the fact that the car could go out of control and make an impact on another car. For example, if the brakes aren’t working you cannot stop the vehicle in time and if the steering wheels went you can’t move the vehicle.

3 0

1 year ago