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sertanlavr [38]
3 years ago
13

What is the maximum mass of sugar that could be dissolved in 1.3 l of water?

Chemistry
1 answer:
Alexeev081 [22]3 years ago
6 0
Answer is: <span>the maximum mass of sugar that could be dissolved in 1.3 l of water in room temperature is 2652 grams.
</span><span>The solubility of sugar in water in room temperature is 204 g/100 ml.
Make proportion, if 204 grams of sugar dissolve in 100 ml of water, than how much dissolve in 1300 ml of water:
204 g : 100 ml = m(sugar) : 1300 ml.
m(sugar) = 204 g </span>· 1300 ml ÷ 100 ml.
m(sugar) = 2652 g ÷ 1000 g/kg = 2,652 kg.
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How much heat is absorbed when 90.5 g of ice is heated from -11.0 °C to 145.0 °C?
Nadusha1986 [10]

Answer:

Q(total) = 283Kj

Explanation:

5 Heat Transitions …

Specific Heats => c(s) = 0.50cal/g∙⁰C,  c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C

Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g

Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.

Q = mcΔT for temperature change regions of the heating curve (single phase only)

Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)

Solid (ice) => Melting Pt  => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal

Melting (s/l) => Liquid (water) =>   Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal

Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal

Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal

Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal

Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)  

                                 = 478cal +7240cal + 9050 cal + 48,870cal + 2036cal

                                 = 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj

4 0
4 years ago
What is the product of the unbalanced equation below?
inessss [21]

Answer:

C. HCI(g)

Explanation:

The following equation between hydrogen gas (H2) and oxygen gas (O2) is given below:

H2(g) + Cl2(g) ►

Based on these unbalanced equation, the products of the reaction was not given, however, if one molecule of hydrogen and oxygen combine, hydrogen chloride (HCl) should be produced as the product of the reaction as in:

H2(g) + Cl2(9) ► 2HCl(g)

4 0
3 years ago
Which statement best describes the compressibility of a gas?
maria [59]

The answer to the question stated above is:
<span> Gas is easily compressible because the molecules of a gas are much further apart than those of a solid.</span>


characteristic properties of gases:

(1) they are easy to compress,

(2) they expand to fill their containers, and

(3) they occupy far more space than the liquids or solids from which they form.

7 0
3 years ago
Lridium-192 is an isotope of iridium and has a half-life of 73.83 days. if a laboratory experiment begins with 100 grams of irid
fiasKO [112]
Radioactive material undergoes first order dissociation kinetics.

For 1st order system,
k = 0.693 / t1/2
where, t 1/2 = half-life of the radioactive disintegration process.

Given that, t 1/2  = <span>73.83 days
Therefore, k = 0.009386 day-1

Also, for 1st order reaction,
k = </span>\frac{2.303}{t} log \frac{Co}{Ct}

Given that, Co = initial concentration of <span>Iridium-192 = 100 g

Therefore, </span>0.009386 = \frac{2.303}{t} log \frac{100}{Ct}

On rearranging we get, Ct = 100 (0.990656)^{t}

Answer: Ct = 100 (0.990656)^{t} equation approximates the amount of Iridium-192 present after t days
8 0
3 years ago
1125 J of energy is used to heat 250 g of iron to 55 °C. The specific heat capacity of iron is 0.45 J/(g·°C).
Stolb23 [73]

Answer:

45 °C.

Explanation:

From the question given above, the following data were obtained:

Heat (Q) = 1125 J

Mass (M) = 250 g

Final temperature (T₂) = 55 °C

Specific heat capacity (C) = 0.45 J/gºC

Initial temperature (T₁) =?

The initial temperature of the iron can be obtained as illustrated below:

Q = MC(T₂ – T₁)

1125 = 250 × 0.45 (55 – T₁)

1125 = 112.5 (55 – T₁)

Divide both side by 112.5

1125/112.5 = 55 – T₁

10 = 55 – T₁

Collect like terms

10 – 55 = –T₁

–45 = –T₁

Multiply through by –1

45 = T₁

T₁ = 45 °C

Therefore, the initial temperature of the iron is 45 °C

8 0
3 years ago
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