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11 days ago

11. If there are 8.24 x 1022 molecules of NaCl in a salt shaker, what is the mass of the salt?

1 answer:

6 0

To determine the mass of salt using Avogadro's number, we find the moles of NaCl: 8.24x10²² molecules NaCl divided by 6.022x10²³ molecules NaCl per mole gives 0.14 mole NaCl. We can convert moles to grams of NaCl by multiplying 0.14 mole by 58g NaCl per mole, yielding a total of 8.12 g NaCl.

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Sodium only has one naturally occuring isotope, 23 Na , with a relative atomic mass of 22.9898 u . A synthetic, radioactive isot

KiRa [2678]

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

6 0

1 month ago

The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO

lorasvet [2504]

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

$87.54 mol O2 x \frac{15}{85}$ = 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2

4 0

11 days ago

Read 2 more answers

An average copper penny minted in the 1960s contained about 3.000 g of copper. how much chalcopyrite had be mined to produce 100

lorasvet [2504]

To find the answer, start by calculating the total mass of the copper utilized:

Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu

Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:

1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.

Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu

= 4.72 moles CuFeS2

The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole CuFeS2 = 866.49 g CuFeS2

8 0

22 days ago

Read 2 more answers

If 3.8 kcal is released by combustion of each gram of glucose, how many kilocalories are released by the combustion of 1.50 mol

castortr0y [2720]

684 kcal. One mole of glucose weighs roughly 180g. Given that 1g of glucose releases 3.8 kcal, we calculate for 1 mole of glucose: 180g -> 180g * 3.8 kcal/g = 684 kcal.

7 0

15 days ago

Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

KiRa [2678]

Answer:NH₃/NH₄Cl

Explanation:

The pH of a buffer can be determined using Henderson-Hasselbalch's equation.

$pH=pKa+log\frac{[base]}{[acid]}$

When the concentration of acid equals that of the base, the pH aligns with the pKa of the buffer. The ideal pH range is pKa ± 1.

Below are the buffers and their corresponding pKa values:

CH₃COONa/CH3COOH (pKa = 4.74)

NH₃/NH₄Cl (pKa = 9.25)

NaOCl/HOCl (pKa = 7.49)

NaNO₂/HNO₂ (pKa = 3.35)

NaCl/HCl Not a buffer

Thus, the ideal buffer is NH₃/NH₄Cl.

4 0

18 days ago