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omeli [17]
2 years ago
15

In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

eeded in the reaction?
1.) 18 g
2.) 54 g
3.) 72 g
4.) 90 g
Chemistry
1 answer:
bulgar [2K]2 years ago
3 0
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol

18g C*1mol C/12 g C = 1.5 mol C

                                 C +     O2 →                CO2

from reaction       1 mol    1 mol              1 mol
from problem     1.5 mol   1.5 mol         1.5 mol

1.5 mol O2*32 g O2/1 mol O2 = 48 g O2

In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount  of O2  that was added for the reaction (probably people who wrote this problem wanted this)
we need  (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is 
</span><span>2.) 54 g.</span>
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B) a helium nucleus moving at a velocity of 1000 mph

Explanation:

According to the De Broglie relation

λ= h/mv

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m= mass of the body

v= velocity of the body.

As we can see from De Broglie's relation, the wavelength of matter waves depends on its mass and velocity. Hence, a very small mass moving at a very high velocity will have the greatest De Broglie wavelength.

Of all the options given, helium is the smallest matter. A velocity of 1000mph is quite high hence it will have the greatest De Broglie wavelength.

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A hydrogen atom is removed from the first carbon atom of a butane molecule and is replaced by a hydroxyl group. draw the new mol
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According to the lab guide, which changes below will you look for in order to test the hypothesis? check all that apply. changes
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Answer:

All of them are.

Explanation:

8 0
2 years ago
Read 2 more answers
(a) At what substrate concentration would an enzyme with a kcat of 30.0 s−1 and a Km of 0.0050 M operate at one-quarter of its m
Dmitrij [34]

The missing graph is in the attachment.

Answer: (a) [S] = 0.0016M

              (b) Vmax = 3V; Vmax = \frac{3V}{2}; Vmax = \frac{11V}{10}

              (c) Enzyme A: black graph; Enzyme B = red graph

Explanation: <u>Enzyme</u> is a protein-based molecule that speed up the rate of a reaction. <u><em>Enzyme</em></u><em> </em><u><em>Kinetics</em></u> studies the reaction rates of it.

The relationship between substrate and rate of reaction is determined by the <u>Michaelis-Menten</u> <u>Equation</u>:

<u />V=\frac{V_{max}[S]}{K_{M}+[S]}<u />

in which:

V is initial velocity of reaction

Vmax is maximum rate of reaction when enzyme's active sites are saturated;

[S] is substrate concentration;

Km is measure of affinity between enzyme and its substrate;

(a) To determine concentration:

0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}<u />

<u />0.25V_{max}(0.005+[S])=V_{max}[S]<u />

<u />0.00125+0.25[S]=[S]<u />

0.75[S] = 0.00125

[S] = 0.0016M

For a Km of 0.005M, substrate's concentration is 0.0016M.

(b) Still using Michaelis-Menten:

<u />V=\frac{V_{max}[S]}{K_{M}+[S]}<u />

Rearraging for Vmax:

V_{max}=\frac{V(K_{M}+[S])}{[S]}

(b-I) for [S] = 1/2Km

V_{max}=\frac{V(K_{M}+0.5K_{M})}{0.5K_{M}}

V_{max}=\frac{V(1.5K_{M})}{0.5K_{M}}

V_{max}= 3V

(b-II) for [S] = 2Km

V_{max}=\frac{V(K_{M}+2K_{M})}{2K_{M}}

V_{max}=\frac{V(3K_M)}{2K_M}

V_{max}=\frac{3V}{2}

(b-III) for [S] = 10Km

V_{max}=\frac{V(K_{M}+10K_M)}{10K_M}

V_{max}=\frac{V(11K_{M})}{10K_{M}}

V_{max}=\frac{11V}{10}

(c) Being the affinity between enzyme and substrate, the lower Km is the less substrate is needed to reach half of maximum velocity.

Km of enzyme A is 2μM and of enzyme B is 0.5μM.

Enzyme B has lower Km than enzyme A, which means the first will need a lower concnetration of substrate to reach half of Vmax.

Analyzing each plot, notice that the red-coloured graph reaches half at a lower concentration, therefore, red-coloured plot is for enzyme B, while black-coloured plot is for enzyme A

<u />

3 0
2 years ago
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

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The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
2 years ago
Read 2 more answers
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