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24 days ago

How many SO32- ions are contained in 99.6 mg of Na2SO3? The molar mass of Na2SO3 is 126.05 g/mol.

2 answers:

7 0

The total number of ions is calculated as follows: Starting with a mass of 99.6 mg, we convert this to grams: 99.6 mg = 0.0996 g. The conversion used is that 1 mg equals 0.001 g. The molar mass of Na2SO3 is 126.05 g/mole. We then determine the number of moles of Na2SO3. Next, we find the quantity of SO32- ions, considering that Na2SO3 dissociates to yield 2 moles of SO32- ions per mole of Na2SO3. Thus, knowing that one mole contains 6.022 x 10^23 ions, this translates to the resulting number of SO32- ions.

castortr0y [2.9K]24 days ago

6 0

99.6 mg = 0.0996 g. Using the conversion factors: 0.0996 g x (1 mol / 125.05 g) x (1 mol SO3 / 1 mol Na2SO3) x (6.02 x 10^22 / 1 mol SO3), the result is 4.79 x 10^19 SO32- ions.

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You want to test the effect of pH on the distribution of Artemia in your 35cm long testing chamber. If you measure a pH = 1 at o

Tems11 [2624]

Answer:The pH measured 10 cm from the most acidic end is 3.42.

Explanation:

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19 days ago

The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its

KiRa [2853]

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

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28 days ago

The ionic radius of a sodium ion is 2.27 angstroms (A) . What is this length in um

lions [2782]

$\boxed{\sf 1Å=10^{-10}m}$

$\\ \rm\longmapsto 2.27Å$

$\\ \rm\longmapsto 2.27\times 10^{-10}m$

$\\ \rm\longmapsto 0.227\times 10^{-9}m$

$\\ \rm\longmapsto 0.0227\times 10^{-8}m$

$\\ \rm\longmapsto 0.00227\times 10^{-7}m$

$\\ \rm\longmapsto 0.00023\times 10^{-6}m$

$\\ \rm\longmapsto 0.00023\mu m$

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1 month ago

If 1.50 μg of CO and 6.80 μg of H2 were added to a reaction vessel, and the reaction went to completion, how many gas particles

Alekssandra [2891]

The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.

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1 month ago

How many molecules are in 13.5g of sulfur dioxide, so2?

alisha [2865]

Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.

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1 month ago