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2 years ago

How many grams of water are frozen if 6,346 Joules of energy was given

Chemistry

1 answer:

ANTONII [103]2 years ago

3 0

Answer:

19g

Explanation:

Given parameters;

Amount of energy = 6346J

Unknown:

Mass of water frozen = ?

Solution:

This is a phase change and a latent heat was involve;

H = mL

H is the quantity of heat

m is the mass

L is the latent heat of fusion of water = 335J/g

So;

6346 = mass x 335

Mass = 19g

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A metal forms the fluoride MF3. Electrolysis of the molten fluoride by a current of 3.86 A for 16.2 minutes deposits 1.25 g of t

larisa86 [58]

Answer: The molar mass of the metal is 96.45 g/mol

Explanation:

The fluoride of the metal formed is $MF_3$

The oxidation half-reaction follows:

$M\rightarrow M^{3+}+3e^-$

Calculating the theoretical mass deposited by using Faraday's law, which is:

$m=\frac{M\times I\times t(s)}{n\times F}$ ......(1)

where,

m = actual mass deposited = 1.25 g

M = molar mass of metal = ?

I = average current = 3.86 A

t = time period in seconds = 16.2 min = 972 s (Conversion factor: 1 min = 60 sec)

n = number of electrons exchanged = $3mol^{-1}$

F = Faraday's constant = 96500 C

Putting values in equation 1, we get:

$1.25g=\frac{M\times 3.86A\times 972s}{3mol^{-1}\times 96500 C}\\\\M=\frac{1.25g\times 3mol^{-1}\times 96500 C}{3.86A\times 972s}\\\\M=96.45g/mol$

Hence, the molar mass of the metal is 96.45 g/mol

7 0

3 years ago

Which two statements about redox reactions are true?

Mars2501 [29]

Answer:

Explanation:

Reduction is a gain of electrons, oxidation is a loss of electrons, and electron transfer reactions are also called redox reactions.

this link might be helpful

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation-Reduction_Reactions

4 0

3 years ago

Read 2 more answers

How does an electron reach an excited state

quester [9]

When an electron quickly occupies an strength state increased than its ground state, it is in an excited state. An electron can end up excited if it is given greater energy, such as if it absorbs a photon, or packet of light, or collides with a close by atom or particle.

5 0

2 years ago

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When 4.15 grams of silver nitrate is reacted with 1.11 grams of iron(III) chloride, which best represents the amount of silver c

PolarNik [594]

Answer:

The mass of silver chloride produced = 2.202 g

Explanation:

Equation of the reaction is given below

3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)

molar mass of AgNO₃ = 170 g/mol

molar mass of FeCl₃ = 233.5 g/mol

molar mass of AgCl = 143.5 g/mol

3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate

4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃

1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃

mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1

therefore, FeCl₃ is the limiting reactant

0.0047 moles of FeCl₃ reacting will produce 0.0047 * 3 moles of AgCl = 0.0141 moles of AgCl

0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =

Therefore mass of silver chloride produced = 2.202 g

3 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

2 years ago