Answer:
34.6 cm³
Explanation:
<em>A chemistry student needs 55.0 g of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of carbon tetrachloride is 1.59 g/cm³. Calculate the volume of carbon tetrachloride the student should pour out. Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of carbon tetrachloride (m): 55.0 g
- Density of carbon tetrachloride (ρ): 1.59 g/cm³
Step 2: Calculate the required volume of carbon tetrachloride
Density is an intrinsic property of matter. It can be calculated as the quotient between the mass of the sample and its volume.
ρ = m/V
V = m/ρ
V = 55.0 g/(1.59 g/cm³)
V = 34.6 cm³
The chemistry student should pour 34.6 cm³ of carbon tetrachloride.
It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. The pathway that produces the most NADH and the least ATP is the KREBS CYCLE. Hope this answers your question. Have a great day ahead!
Answer:
yes
Explanation:
because of high positive charge of Nb6+ delta G is more negative for Nb6+ as an oxidising agent♡
Multiplying by 1000 will allow you to convert 35 liters to milliliters
<h3>Conversion scale</h3>
1 Liters = 1000 milliliters
Thus, multiply the value in liter by 1000 will convert it to milliters
With the above scale, we can convert 35 liters to milliliters. Details below:
<h3>How to convert 35 liters to milliliters</h3>
We can convert 35 liters to milliliters as illustrated below:
1 Liters = 1000 milliliters
Therefore,
35 liters = (35 liters × 1000 milliliters) / 1 liters
35 liters = 35000 milliliters
Thus, 35 liters is equivalent to 35000 milliliters
Learn more about conversion:
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Zn + I2 = ZnI2
Percentage yield = actual yield / theoretical yield * 100
Actual yield of ZnI2 = 515.6g
Mole of zinc reacted = 125 / 65.4 = 1.91 mol
for every 1 mole of zinc will form 1 mole of ZnI2
Mole of ZnI2 formed = 1.91 * 1 = 1.91 mol
Theoretical yield of ZnI2 = 1.91 * (65.4 + 126.9*2) = 609.672 g
515.6 / 609.672 * 100 = 84.6%
