Answer: The answer is A. A conductor that allows electricity to flow easily
Answer:
Glucose
Explanation:
Glucose is broken down into water and carbon dioxide.
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Answer:

Explanation:
<u>1. Convert Molecules to Moles</u>
First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

Multiply by the given number of molecules.

Flip the fraction so the molecules cancel out.




<u>2. Convert Moles to Grams</u>
Next, we convert moles to grams using the molar mass.
We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.
- Na: 22.9897693 g/mol
- O: 15.999 g/mol
- H: 1.008 g/mol
Since the formula has no subscripts, we can simply add the molar masses.
- NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol
Use this as a ratio.

Multiply by the number of moles we calculated.

The moles of sodium hydroxide cancel.



The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.
The concentration in weight percent when 6 g of sugar is mixed with 9 g of water is 40%.
There are several ways to denote the concentration of a solution like
- Molarity
- Molality
- Mass percent
- Mole Fraction
The formula for calculating mass percent is as follows
Mass per cent = (Mass of solute/Mass of solute + Mass of solvent) x 100%
In the given situation sugar is the solute and water is the solvent.
Putting the given values in the above formula
Mass per cent = (6/6+9) x 100%= 6/15 x 100% = 40%
Hence, the concentration in weight percent when 6 g of sugar is mixed with 9 g of water is 40%.
To know more about "concentration of a solution", refer to the link given below:
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Answer:
Rate ≅ 1.01 M/s (3 sig. figs.)
Explanation:
Given A(g) + B(g) => AB(g)
Rate = k[A(g)][B(g)]²
at Rate (1) = 0.239M/s = k[2.00M][2.00M]² => k = (0.239M/s) / (2.00M)(2.00M)²
k = 0.29875 M⁻²·s⁻¹
Rate (2) = k[A(g)][B(g)]² = (0.29875M⁻²·s⁻¹)(4.81M)(2.65M)² = 1.009124472 M/s (calc. ans.) ≅ 1.01 M/s (3 sig. figs.)