How many grams of Br are in 445 g CaBr2 ?

Answer:

The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.

0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3

0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.

Answer:

When you heat ice, its temperature rises, but as soon as the ice starts to melt, the temperature stays constant until all the ice has melted. This happens because all the heat energy goes into breaking the bonds of the ice's crystal lattice structure.

Explanation:

The temperature stop rising while ice melts into water because, the energy is being absorbed to separate the particles. This is because of latent heat of fusion.

The amount of energy needed to convert the solid substance into a liquid substance by modifying its physical effects. It exists also named enthalpy of fusion. When heat exists supplied to ice, it begins melting and heat is used to increase temperature initially. But after the occasional temperature of ice does not vary and the extra heat exists utilized to melt the ice by cracking bonds between crystal lattice of ice.

The temperature stops increasing while the ice melts into the water because the energy exists being absorbed to divide the particles. This exists because of the latent heat of fusion.

Therefore, (D) option is the correct answer.

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Thermochemistry has to do with  heat evolved or absorbed in a chemical reactions. Thermochemical equations are equations in which the heat of reaction is included in the reaction equation. The reaction of moles and heat of reaction is important here.

This question has to do with thermochemistry and thermochemical equations.

The answers to each of the questions are shown below;

a) 300.52 KJ

b) 11.39 g

c) 5.78 g

The equation of the thermochemical reaction is;

2C12H26 + 37O2-------> 24CO2 + 15026KJ

Number of moles of CO2 released = 21.3g/44g/mol = 0.48 moles

From the reaction equation;

15026KJ is released when 24 moles of CO2 is released

x KJ is released when  0.48 moles of CO2 is released

x = 15026KJ  * 0.48 moles/24 moles

x = 300.52 KJ

b) If 2 moles of C12H26 released 15026KJ of heat

     x moles of C12H26  released 500.00KJ

x = 2 * 500.00KJ/15026KJ

x = 0.067 moles

Mass of C12H26 consumed =  0.067 moles * 170 g/mol = 11.39 g

c) Heat gained by water = heat released by combustion of kerosene

Heat gained by water = 0.75 Kg * 4200  * (90 -10)

Heat gained by water = 252 KJ

If 2 moles of C12H26  produced 15026KJ

x moles of C12H26  produces 252 KJ

x = 2 * 252/15026

x = 0.034 moles

Mass of C12H26   = 0.034 moles *  170 g/mol = 5.78 g

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Answer: The mass of copper (II) nitrate produced is 105.04 g.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of copper = 35.5 g

Molar mass of copper = 63.5 g/mol

Plugging values in equation 1:

[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]

The given chemical equation follows:

[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]

By the stoichiometry of the reaction:

If 3 moles of copper produces 3 moles of copper (II) nitrate

So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate

Molar mass of copper (II) nitrate = 187.56 g/mol

Plugging values in equation 1:

[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]

Hence, the mass of copper (II) nitrate produced is 105.04 g.

since we are given the moles for Co and O, we'll divide both of those moles by the lowest mole quantity, which is, in this case, 2.59. After dividing, we see that the ratio of O to Co is 2:1. So, for every 1 Co atom, there has to be 2 O atoms. we can then insert the 2 in for OH to satisfy this ratio.

Answer:

1087.84 J

Explanation:

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

Mass of metal (Mₘ) = 70 g

Temperature of metal (Tₘ) = 80 °C

Mass of water (Mᵥᵥ) = 100 g

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) = 24.6 °C

Heat lost by metal (Qₘ) =?

NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Thus, we shall determine the heat gained by water. This can be obtained as follow:

Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

Qᵥᵥ = 100 × 4.184 (24.6 – 22)

Qᵥᵥ = 418.4 × 2.6

Qᵥᵥ = 1087.84 J

Thus, the heat gained by water is 1087.84 J.

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Qᵥᵥ = 1087.84 J

Qₘ = 1087.84 J

Therefore, the heat lost by the metal is 1087.84 J

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.

Let's use the following expression to calculate the heat absorbed by the water.

Qw = c × m × ΔT

Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ

where,

According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.

Qw + Qm = 0

Qm = -Qw = -10.8 kJ

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

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Answer:

Average specific heat capacity of metal = 0.57 J/g°C

Explanation:

Heat lost = Heat gained

Heat energy gained or lost, H = mcΔT

where m = mass of substance, c = specific heat capacity, ΔT = temperature change

Trial 1:

Heat lost by metal = -[2.746 g × c × ΔT]

ΔT = (26.3 - 72.1) °C = -45.8 °C

Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C

Heat gained by water = 15.200 × 4.18 × ΔT

ΔT = (26.3 - 24.7) = 1.6 °C

Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J

From Heat lost = Heat gained

c × (125.7688)g°C = 101.6576 J

c = 101.6576 J / 125.7688 g°C

c = 0.8083 J/g°C

Trial 2:

Heat lost by metal = -[2.750 g × c × ΔT]

ΔT = (26.2 - 72.2)°C] = - 46 °C

Heat lost by metal = -[2.750 g × c × (-46 °C)

Heat lost by metal = c × (126.5) g°C

Heat gained by water = 15.206 × 4.18 × ΔT

ΔT = (26.2 - 24.6) = 1.6 °C

Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J

From Heat lost = Heat gained

c × (126.5)g°C = 101.6977 J

c = 101.697728 J / 126.5 g°C

c = 0.8039 J/g°C

Trial 3:

Heat lost by metal = -[2.900 g × c × ΔT]

ΔT = (24.7 - 71.9)°C] = - 47.2 °C

Heat lost by metal = -[2.900 g × c × (- 47.2 °C)

Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C

Heat gained by water = 15.201 × 4.18 × ΔT

ΔT = (24.7 - 24.5) = 0.2 °C

Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J

From Heat lost = Heat gained

c × (136.88)g°C = 12.708036 J

c = 12.708036 J / 136.88 g°C

c = 0.0928 J/g°C

Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3

Average specific heat capacity of metal = 0.57 J/g°C

Answer:

Fe(NO3)3, Cr(NO3)3, Co(NO3)3

Explanation:

According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.

Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.

The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.

The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.

Answer:

main asteroid belt separates the inner planets from the outer planets

Answer: The property of magnesium that is exhibited by it is DUCTILITY. The correct option is A.

Explanation:

Magnesium is a member of the alkaline earth metals. It occurs in nature, only in the combined state, as Epsom salt, dolomite and in many trioxosilicates( IV) including talc and asbestos. They have the following physical properties:

--> Appearance: they are silvery-white solids

--> Relative density: It has a relative density of 1.74

--> DUCTILITY: it's very ductile in nature

--> melting point: it has a melting point of 660°C.

--> Conductivity: They are good conductor of heat and electricity.

Furthermore, DUCTILITY is the physical property of a metal associated with the ability to be hammered thin or stretched into wire without breaking. A metal such as magnesium can therefore be coiled as a thin ribbon without fracturing due to its ductile physical properties.

Answer:

C

Explanation:

b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction

[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex] is the redox reaction. Hence, option C is correct.

A chemical reaction taking place between an oxidizing substance and a reducing substance.

The oxidizing substance is used to lose electrons in the reaction, and the reducing substance is used to gain electrons.

On the reduction potential chart, zinc is a stronger oxidizing agent than, Copper (Cu), which is a reducing agent as compared to silver

The redox reaction most likely occurs if silver and copper metal were added to a solution that contained silver and copper ions is ;

[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex]

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Answer:

Two molecules of hydrogen combine with two molecules of oxygen to form hydrogen peroxide. Hence, its chemical formula is H2O2. It is the simplest peroxide (since it is a compound with an oxygen-oxygen single bond). Hydrogen peroxide has basic uses as an oxidizer, bleaching agent and antiseptic

Answer:

c

Explanation:

table of data help us to understand and present our work better

I did it and got it right, it's c

Answer:

endet nach selam nw

4gh7

Answer:

6.684g

Explanation:

Here, we can use the mole ratio of the gases to calculate.

We know that the mole ratio of the gases equate to their number of moles.

Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol

Thus, the number of moles produced is 5.98/32 = 0.186875

Where do we move from here?

We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.

449/851 = 0.186875/n

n =(0.186875 * 851)/449

n = 0.3542

Now we do the same for argon to get the number of moles of argon.

Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.

P = 851 - 449 = 402 mmHg

We now use the mole ratio relation.

402/851 = n/0.3542

n = (402 * 0.3542) / 851

n = 0.1673

Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.

The atomic mass of argon is 39.948 amu

The mass is thus 39.948 * 0.1673 = 6.684g

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

Answer:

Line graphs are used to track changes over short and long periods of time. When smaller changes exist, line graphs are better to use than bar graphs. Line graphs can also be used to compare changes over the same period of time for more than one group.

Answer:

7.38*10^21

Explanation:

2Na+2H20=2NaOH+H2

nNa=0.0123

number of water moles: 0.012*6*10^23=7.38*10^21

The question is incomplete, the complete question is;

When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.

First, the oxidant, like hypochlorite, oxidizes

Choose...

neutral iodine into iodide ion

iodide ion into neutral iodine

iodate polyatomic ion into iodide ion

When starch and iodine are both present, the solution is

Choose...

blue-black

brownish yellow

clear

During the titration, the titrant, like thiosulfate, reduces the

Choose...

iodide ion into iodate polyatomic ion

neutral iodine into iodide ion

iodide ion into neutral iodine

When the iodine has completely reacted at the endpoint of the titration, the solution should become

Choose...

clear

blue-black

brownish yellow

Answer:

1. iodide ion into neutral iodine

2. blue-black

3. neutral iodine into iodide ion

4. clear

Explanation:

Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;

ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)

When iodine is added, the colour of the starch solution immediately changes to blue-black.

A reduction reaction occurs when the titrant, thiosulfate is added as follows;

I2 + 2S2O32- → 2I- + S4O62-

The solution at end point is found to become clear again.

Answer:

[tex]pH=4.77[/tex]

Explanation:

From the question we are told that:

pKa for Acetic Acid [tex]pK_a= 4.77[/tex]

Therefore

For Equal Concentration of acetic acid and acetatic ion

[tex]CH_3COOH=CH_3COO^-[/tex]

Generally the Henderson's equation for pH value is mathematically given by

[tex]pH=pK_a+log\frac{base}{acid}[/tex]

[tex]pH=4.77+log\frac{CH_3COO^-}{CH_3COOH}[/tex]

[tex]pH=4.77+log1[/tex]

[tex]pH=4.77[/tex]

Answer:

46.2%

Explanation:

Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles

Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.

Hence;

Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g

% yield = actual yield/theoretical yield × 100

% yield = 1.7 g/3.68 g × 100

% yield = 46.2%

Answer:

4.27 g

Explanation:

Number of moles = concentration × volume

Concentration = 0.121 M

Volume = 250 mL

Number of moles = 0.121 M × 250/1000 L

Number of moles = 0.03 moles

Number of moles = mass/molar mass

Mass= Number of moles × molar mass

Mass= 0.03 moles × 142.394 g/mol

Mass = 4.27 g

Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

[tex]M=58g[/tex]

Explanation:

From the question we are told that:

Heat Capacity [tex]H=0.897[/tex]

Mass of water [tex]M=200g[/tex]

Initial Temperature of Aluminium [tex]T_a=85.6[/tex]

Initial Temperature of Water [tex]T_{w1}=16.0[/tex]

Final Temperature of Water  [tex]T_{w2}=16.0[/tex]

Generally

Heat loss=Heat Gain

Therefore

[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]

[tex]M=58g[/tex]

Answer:

D

Explanation:

Desalination

Removing salt from sea water is known as desalination

Explanation:

the molecular weight of K2SO3 is 158. 2598 m/s.

Answer:

C

Explanation:

if we were to followw the IUPAC

Answer:

information is missing

Explanation

reaction is needed to solve the problem

Explanation:

here's the answer to your question

Kc require (aqueous/gaseous) products to be on the numerator and (aqueous/gaseous) reactants to be in the denominator, whereas Ksp will require (aqueous) products to be on the numerator and (aqueous) reactants to be in the denominator. Both require products on top and reactants in the bottom.

K = [products] / [reactants]

Kc is used when a reaction reaches dynamic equilibrium, whereas Ksp is used when an insoluble ionic solid dissolved by a tiny amount in a solution, as well as in determining whether or not a precipitate will form.

Kc can be used to measure equilibrium concentration for all reactions, whereas Ksp is limited to only ionic compounds' solubility.

The decomposition of HI (g) will required the use of Kc since the species are all gaseous, and gases cannot be ionic.