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2 years ago

Match the following terms:

Chemistry

1 answer:

AURORKA [14]2 years ago

3 0

Answer:

See below ~

Explanation:

A to B

⇒ Solid Phase

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B to C

⇒ Melting

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C to D

⇒ Liquid Phase

==========================================================

D to E

⇒ Vaporizing

============================================================

E to F and beyond

⇒ Gas Phase

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What is the percent composition of N in HNO3 ?

bonufazy [111]

Answer:

22.22%

Explanation:

N/HNO3*100

14/63*100

=22.22%

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3 years ago

Which method would you use to separate silt from water

jonny [76]

The answer is "filtration"

Filtration is used to seprate a solid from a liquid in which it is suspended. Filtration is also used to separate a substance from a mixture because one is insoluble in the solvent and the other is solube. The separation is due to particle size.

I hope this helps.

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3 years ago

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A solution is prepared by dissolving ammonium sulfate in enough water to make of stock solution. A sample of this stock solution

Pani-rosa [81]

Answer: molarity of ammonium ions = 0.274mol/L

molarity of sulfate ions = 0.137mol/L

Note: The complete question is given below

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Explanation:

Molar concentration = no of moles/volume in liters

no of moles = mass/molar mass

mass of ammonium sulfate = 10.8g, molar mass of ammonium sulfate, (NH₄)₂SO₄ = (14+4)*2 + 32+ (16)*4 = 132g/mol

no of moles = 10.8g/132g/mol = 0.0820moles

Molarity of stock solution = 0.0820mol/(100ml/1000ml* 1L) = 0.0820mol/0.1L Molarity of stock solution = 0.820mol/L

Concentration of final solution is obtained from the dilution formula,

C1V1 = C2V2

C1 = 0.820M, V1 = 10mL, C2 = ?, V2 = 60mL

C2 = C1V1/V2

C2 = 0.820*10/60 = 0.137mol/L

molar concentration of ions = molarity of solution * no of ions

molarity of ammonium ions = 0.137mol/L * 2 = 0.274mol/L

molarity of sulfate ions = 0.137 mol/L * 1 = 0.137mol/L

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3 years ago

How is geothermal produced

soldier1979 [14.2K]

The steam rotates a turbine that activates a generator, which produces electricity. Many power plants still use fossil fuels to boil water for steam. Geothermal power plants, however, use steam produced from reservoirs of hot water found a couple of miles or more below the Earth's surface.

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3 years ago

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does not go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago