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larisa [96]
3 years ago
15

Calculate the mole fraction of the total ions in an aqueous solution prepared by dissolving 0.400 moles of CaI2 in 850.0 g of wa

ter.
Chemistry
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

0.00840

Explanation:

The computation of the mole fraction is as follow:

As we know that

Molar mass = Number of grams ÷ number of moles

Or

number of moles = Number of grams ÷ molar mass

Given that

Number of moles of CaI2 = 0.400

And, Molar mass of water = 18.0 g/mol

Now  Number of moles of water is

= 850.0 g ÷ 18.0 g/mol

= 47.22 mol

And, Total number of moles is

= 0.400 + 47.22

= 47.62

So, Molar fraction of CaI2 is

= 0.400 ÷ 47.62

= 0.00840

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What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?
podryga [215]

Answer:

17.934 kg of water

Explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂  →  (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) →  8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then  1.2 x 10²⁶ molecules will give:

\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times  10^{23}}

= 17.934 x 10³ g of water

= 17.934 kg of water

5 0
3 years ago
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
3 years ago
Determine the possible traits of the calves if:
White raven [17]

Answer:

1. All red calves i.e. RR

2. All roan calves i.e RW

3. 2 red calves (RR) and two roan calves (RW)

Explanation:

According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):

1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR

2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW

3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).

7 0
3 years ago
Anyone know this please help me
natka813 [3]

Answer:go to the sight called would or do help

Explanation:go onnnnnnn

6 0
3 years ago
Protozoan infection is not likely to be transmitted by:
anygoal [31]
Hi! 

The answer is breathing.

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Breathing
is not considered to be likely to transmit protozoan infection because protozoans rarely develop in the respiratory tract.
6 0
3 years ago
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