How many chloride ions are in a 220 grams of calcium chloride?

You have 125 g of a certain seasoning and are told that it contains 62.0 g of salt. what is the percentage of salt by mass in th

KiRa [2933]

62.0g divided by 125g equals 0.496, then multiplied by 100 gives 49.6%.

4 0

2 months ago

How many grams of AlF3 are in 2.64 moles of AlF3?

Anarel [2989]

To determine the mass of AlF3 in 2.64 moles of AlF3, we use the formula: mass = moles x molar mass, which results in 221.76 grams of AlF3.

3 0

2 months ago

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as

Tems11 [2777]

The question lacks completeness; the full question is:

Determine the theoretical yield:

When excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide, sodium nitrate and copper(II) sulfide precipitate. For this reaction, 469 grams of copper(II) nitrate was combined with 156 grams of sodium sulfide yielding 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

$Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)$

Calculating moles of copper(II) nitrate = $\frac{469 g}{187.5 g/mol}=2.5013 mol$

Moles of sodium sulfide = $\frac{156 g}{78 g/mol}=2 mol$

Based on the reaction, 1 mole of copper(II) nitrate reacts with 1 mole of sodium sulfide.

Thus, 2 moles of sodium sulfide will react with:

$\frac{1}{1}\times 2mol= 2 mol$ of copper(II) nitrate

Since sodium sulfide is in limiting quantities, the amount of sodium nitrate produced will depend on the moles of sodium sulfide available.

According to the reaction, 1 mole of sodium sulfide generates 2 moles of sodium nitrate; thus, 2 moles of sodium sulfide will yield:

$\frac{2}{1}\times 2mol=4 mol$ sodium nitrate

The total mass of 4 moles of sodium nitrate is:

85 g/mol × 4 mol = 340 g

The theoretical yield of sodium nitrate amounts to 340 g.

The theoretical yield of sodium nitrate is 340 grams.

7 0

1 month ago

Read 2 more answers

A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

lorasvet [2795]

Answer:

22.7

Explanation:

Initially, calculate the energy released from the sample's mass. The combustion heat represents the energy per mole of the fuel:

ΔHC=qrxnn

We can rearrange this formula to isolate qrxn, remembering to convert the sample's mass into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released during the reaction must match the total heat absorbed by both the water and the bomb calorimeter:

qrxn=−(qwater+qbomb)

The heat absorption by the water can be calculated using its specific heat:

qwater=mcΔT

The calorimeter's heat absorption can be derived from its heat capacity:

qbomb=CΔT

Combine both equations into the first equation, substituting the known details, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute each multiplication term and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)−(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)−(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)−50208 J+(420. J∘C×Tfinal)−8400 J]

Combine the similar terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, isolate Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

Round to three significant figures gives the final result as 22.7∘C.

8 0

2 months ago