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tekilochka [14]
3 years ago
9

⦁ How many moles of carbon dioxide will be produced by the complete reaction of 2.0 mol of glucose (C6H12O6), according to the f

ollowing equation?
C6H12O6 + 6O2 → 6CO2 + 6H2O
Chemistry
1 answer:
SashulF [63]3 years ago
3 0

Answer:

i think 12 moles of CO2 will be produced

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Eduardwww [97]

Answer:

The answer is C, number 4 shows a rift valley

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7 0
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If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most
andrey2020 [161]

Answer:

c

Explanation:

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3 years ago
What is the percentage error of length measurement of 0.229cm if the correct value is 0.225cm
Mars2501 [29]

Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.

So first, we take our measured value, .299 cm, minus our known value, .225 cm.

.299 cm - .225 cm=.004 cm

Next, we divide that by our known value

\frac{.004}{.225}=.0177777778

Finally, multiply your answer by 100

.0177777778 x 100= 1.77777778 %

Round to three significant figures, and you're done.

=1.78 % error



5 0
3 years ago
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Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the sol
Afina-wow [57]

Answer:

1.811 g

Explanation:

The computation of the mass need to use to make the solution is shown below:

We know that molarity is

Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}

So,

Number\ of\ moles = Molarity\ \times Volume\ in\ L

= 0.223\times 0.141

= 0.031 moles

Now

Mass = moles \times Molecualr\ weight

where,

The Molecular weight of NaCl is 58.44 g/mole

And, the moles are  0.031 moles

So, the mass of NaCL is

= 0.031 \times 58.44

= 1.811 g

We simply applied the above formulas

3 0
3 years ago
fictitious element Jz has two common isotopes. Three-fifth of the isotopes have a mass of 329.1 amu, while the rest have a mass
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The average atomic mass of Jz is 331.3 u.

The average atomic mass of Jz is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its fractional abundance).  

Thus,

⅗ × 329.1 u = 197.46  u

⅖ × 334.5 u = <u>133.80 u</u>

      TOTAL = 331.3   u

7 0
3 years ago
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