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OLEGan [10]
3 years ago
8

You contain the same elements that are found in and around the Earth. The atmosphere contains mostly nitrogen, 78%, and oxygen,

21%. These two elements chemically combine with carbon and hydrogen to form an important compound found in all living things. What is that compound?
A) carbohydrates
B) lipids
C) proteins
D) sugars
Chemistry
1 answer:
Alexandra [31]3 years ago
5 0

Answer:

protein and lipids

Explanation:

Proteins, like carbohydrates and lipids, are compounds of carbon hydrogen and oxygen. However, unlike carbohydrates and lipids, proteins also contain the element nitrogen and are hence referred to as nitrogenous compounds. In addition to nitrogen, proteins may contain phosphorus or sulphur or both. Some proteins such as haemoglobin contain other elements such as iron.

If this helps please mark me as brainly, if you don't want to that's ok. im just here to help you get good grades :D

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Explain specifically how an electron gives off light in an atom.
exis [7]

Answer:

Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.

Explanation:

When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.

3 0
2 years ago
How many moles of xenon gas, Xe, would occopy 37.8L at stp
Charra [1.4K]

Answer:

The number of moles of  xenon are 1.69 mol.

Explanation:

Given data:

Number of moles of xenon = ?

Volume of gas = 37.8 L

Temperature = 273 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values in formula.

1 atm × 37.8 L = n ×  0.0821 atm.L/ mol.K   ×273 K

37.8 atm.L =  n × 22.413 atm.L/ mol.

n = 37.8 atm.L /  22.413 atm.L/ mol.

n = 1.69 mol

The number of moles of  xenon are 1.69.

8 0
3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
David kept 100 grams of radioactive sodium in a container. He observed the amount of sodium left in the container after regular
Anna [14]
First we have to understand what a half life is. A half life is a measure of time when the amount of a certain object is 50% of the original amount. Hence the answer for this is letter A. The initial amount is 100. fifty percent of 100 is 50 and that happens after 15 hours. Hence, 15 hours is the half life period. 
5 0
3 years ago
Read 2 more answers
A sample of nitrogen gas is stored in a 489.6 mL flask at 108 kPa and 10.0°C. The gas is transferred to a 750.0 mL flask at 28.7
Wewaii [24]

Answer:

P2= 125.26 Kpa

Explanation:

V1= 489.6 ml=0.4896L

V2= 750 ml= 0.750L

V1= 180 KPa= 180000 Pa

P2= ?

T1= 10 = 10 + 273.15 = 283.15K

T2= 28.7+273.15= 301.85K

180000Pa* 0.4896L/ 283.15K * 301.85K/0.75L

P2= 12526.28553

P2= 125.26 KPa

5 0
3 years ago
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