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Scrat [10]
3 years ago
11

Why does water float on corn syrup

Chemistry
2 answers:
NISA [10]3 years ago
5 0

Answer:

density.

Explanation:

A cubic centimeter of water weighs 1 gram. Since a cubic centimeter of vegetable oil weighs less than 1 gram, oil will float on water. Corn syrup is more dense than water so 1 cubic centimeter of corn syrup weighs more than 1 gram. Therefore, corn syrup sinks in water.

Mrac [35]3 years ago
3 0

Answer:

Click on the photo to see the full attachment

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How many molecules are contained in 0.800 mol o2?
Ksenya-84 [330]
Multiply .800 moles of O2 by Avagadro's number divided by 1 mole. This will get rid of the moles on the bottom and leave you with molecules. So technically .800 times 6.02x10^23.
6 0
2 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
Which of the following is an example of kinetic energy?
deff fn [24]

Answer:

A. The energy of moving particles near a fire

Explanation:

7 0
3 years ago
A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai
marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

  • C: 12.011;
  • O: 15.999.

Calculate the molar mass of carbon dioxide \rm CO_2:

M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

Find the number of moles of molecules in that 41.1\;\rm g sample of \rm CO_2:

n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol.

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

P \cdot V = n \cdot R \cdot T,

where

  • P is the pressure inside the container.
  • V is the volume of the container.
  • n is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
  • R is the ideal gas constant.
  • T is the absolute temperature of the gas.

Rearrange the equation to find an expression for P, the pressure inside the container.

\displaystyle P = \frac{n \cdot R \cdot T}{V}.

Look up the ideal gas constant in the appropriate units.

R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}.

Evaluate the expression for P:

\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}.

Apply dimensional analysis to verify the unit of pressure.

4 0
3 years ago
If a solution containing
tankabanditka [31]

Answer:

Hg(NO₃)₂(aq) + Na₂SO₄(aq)   →   2NaNO₃(aq) + HgSO₄(s)

Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

                                        = 0.1172 x 296.65

                                        = 34.757g

Explanation:

6 0
3 years ago
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