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vodomira [7]
2 years ago
15

Sugars such as glucose, fructose, and ribose are examples of _______. A. proteins B. nucleic acids C. carbohydrates D. lipids

Chemistry
2 answers:
algol132 years ago
8 0
Carbonhydrates so it is c

Eddi Din [679]2 years ago
7 0
The correct answer is C, carbohydrates. 
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How many number of oxygen molecules are required to produce 0.22 g of CO2 according to the reaction?
zmey [24]
The number of moles for co2=mass(g)/molar mass
n=.22/44=.005 mole of CO2
from the equation we see the relationship between the moles of co2 and O2 and we find that they have the same number of moles
So we need .005mole of O2
Multiple the number of moles with avogadro’s number to know the number of O2molecules
.005x6.022 x10^23
3 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r
kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)

Energy absorbed by pork,E = 1.6\times 10^3 kJ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

10\%=\frac{E}{Q}\times 100

Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = -1.6\times 10^4 kJ

Moles of propane burnt to produce Q energy =n

n\times \Delta H_{rxn}=Q

n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

\frac{1}{3}\times 7.83 mol=23.49 mol carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0
2 years ago
Confused as heck. please help!
Nitella [24]

Answer: that all thre water cycle and C is vaporation

Explanation:

6 0
2 years ago
(((NEED ANSWER QUICK!!!)))<br><br> Which is the stronger conjugate base, CN- or OCN-? Explain
Shkiper50 [21]

Answer:

The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.

Explanation:

Given conjugate base CN⁻ => weak acid => HCN =>  Ka =4.9 x 10⁻¹⁰

Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴

Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻

6 0
3 years ago
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