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3 years ago

Which of the following is not a simple machine

Chemistry

1 answer:

dedylja [7]3 years ago

5 0

C: screw I believe so

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Plz answer 3 ,4,5 thanks

ratelena [41]

For #3 the answer is B, for #4 the answer is F, and for #5 the answer is B.

5 0

3 years ago

The electrons stripped from glucose in cellular respiration end up in which compound? The electrons stripped from glucose in cel

Airida [17]

The final destination to where some of the electrons go to at the end of cellular respiration would be D. Oxygen. Assuming that this aerobic cellular respiration, the final electron acceptor is that of oxygen.

6 0

3 years ago

The element that doesn't occur in elemental state in nature is

Pepsi [2]

Answer:

technetium, atomic number 43; promethium, number 61; astatine, number 85; francium, number 87; neptunium, number 93; and plutonium, number 94.

Explanation:

4 0

2 years ago

The Prandtl number, Npr, is a dimensionless group important in heat transfer calculations. It defined as Cuk, where C, is the he

VARVARA [1.3K]

Answer:

The Prandlt number for this fluid is 1630.

Explanation:

The Prandlt number is defined as:

$Pr=\frac{C_p\mu}{k}$

To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.

$\mu=1936\frac{lb}{ft*h}*\frac{1000g}{2.205lb}*\frac{3.281ft}{1m}*\frac{1h}{3600s} \\ \\\mu=800 \frac{g}{m*s}$

Now that we have coherent units, we can calculate Pr

$Pr=\frac{C_p\mu}{k}=\frac{0.583*800}{0.286}= 1630$

8 0

3 years ago

A reaction is followed and found to have a rate constant of 3.36 × 104 m-1s-1 at 344 k and a rate constant of 7.69 m-1s-1 at 219

g100num [7]

The relation between rate constants at different temperatures, temperature and activation energy is known as Arrhenius equation

<u>Arrhenius equation</u>

$K=Ae^-{\frac{Ea}{RT} }$

For two temperatures

$Ea=R(\frac{ln\frac{k1}{k1} }{(\frac{1}{T1})-(\frac{1}{T2})})$

Where

Ea = ? = activation energy

k1 = 3.36 × 10⁴

T1=344 k

k2=7.69

T2=219K

R= gas constant = 8.314 J /molK

Putting values

$Ea= (8.314)(\frac{ln(\frac{7.69}{33600})}{(\frac{1}{344})(\frac{1}{219})}$

Ea = (-69.69)/(-0.00166) = 41981.93 J/mol

Activation energy is 42.0 kJ /mol

3 0

3 years ago