What types of reactions are these 3 chemical equations?-
*2Kl+Pb(NO3)2=PbI2+2KNO3
*2Al+3CuSO4=Al2(SO4)3+3Cu
*C2H5OH+3O2=2CO2+3H2O

Answer:

The answer is C

Explanation:

The more space the molecules have the faster they will move, solid doesn't allow movement at all, when it gets to liquid they are free to move around because it's more space, when a gas they can move all around in the air.

Explanation:

2H2O and 0.722g of Na3PO4.12H20 mix with water forming 500 mL of solution, how many grams of Ba3(. ... Forming 500 ML Of Solution, How Many Grams Of Ba3(PO4)2 Precipitate?

When 0.421 g of BaCl₂ · 2 H2O and 0.722g of Na₃PO₄ · 12 H₂O mix with water forming 500 mL of solution then 0.003 mol/L of Ba₃(PO4)₂ precipitate.

Stoichiometry is an important concept which helps us use the balanced chemical equation to measures the amount of reactants and products.

Molarity of solution is the number of moles of solute per liter of solution. Molarity is also called Molar Concentration.

Molarity (M) = [tex]\frac{\text{number of moles}}{\text{volume of solution (in liters)}}[/tex]

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

First we have to write the balanced chemical equation

2Na₃PO₄ · 12 H₂O + 3BaCl₂ · 2 H2O → Ba₃(PO4)₂ + 6 NaCl + 30 H₂O

Here 3 moles of BaCl₂ · 2 H2O produce 1 mole of Ba₃(PO4)₂

So,

Number of moles of BaCl₂ · 2 H2O = [tex]\frac{\text{Mass of}\ BaCl_2 . 2H_2O}{\text{Molar mass of}\ BaCl_2. 2H_2O}[/tex]

                                                           =  [tex]\frac{0.421}{244.26}[/tex]

                                                           = 0.017

Number of moles of 2Na₃PO₄ · 12 H₂O = [tex]\frac{\text{Mass of}\ Na_3PO_4. 12H_2O}{\text{Molar mass of}\ Na_3PO_4 . 12H_2O}[/tex]

                                                                 = [tex]\frac{0.722}{380.12}[/tex]

                                                                 = 0.002

Therefore, Na₃PO₄ · 12 H₂O is the limiting reagent.

Now, put the value in above formula to find the molarity

Molarity (M) = [tex]\frac{\text{number of moles}}{\text{volume of solution (in liters)}}[/tex]

                    = [tex]\frac{0.0017}{0.5}[/tex]

                    = 0.003 mol/L

Thus, from above conclusion we can say that 0.003 mol/L of Ba₃(PO4)₂ precipitate.

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Answer:

I think you forgot to post the question or picture

Answer:

28.55 mL or 0.02855 L

Explanation:

Applying,

Density(D) = mass(m)/Volume(V)

D = m/V............... Equation 1

Where D = Density of the Bromine, m = mass of the bromine, V = volume of the bromine

make V the subeject of the equation

V = m/D............. Equation 2

From the question,

Given: m = 88.5 g, D = 3.10 g/mL

Substitute these values into equation 2

V = 88.5/3.10

V = 28.55 mL

Hence the volume occuppied by the Bromine gas is  28.55 mL or 0.02855 L

Answer:

Option A. The specific heat is 0.897 J/gºC, The Substance is aluminum.

Explanation:

We'll begin by calculating the change in temperature of the substance. This can be obtained as follow:

Initial temperature (T₁) = 20.0 °C

Final temperature (T₂) = 40.0 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 40 – 20

ΔT = 20 °C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Mass (M) = 136 g

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 2440 J

Specific heat capacity (C) =?

Q = MCΔT

2440 = 136 × C × 20

2440 = 2720 × C

Divide both side by 2720

C = 2440 / 2720

C = 0.897 J/gºC

Comparing the specific heat capacity (i.e 0.897 J/gºC) of the substance with those in the table above, the substance is Aluminum.

Thus, option A gives the correct answer to the question.

Answer:

[tex]1 \: l \: contains \: 0.20 \: moles \: of \: HOCN \\ 0.03 \: l \: will \: contain \: (0.03 \times 0.20) \: moles \\ = 0.006 \: moles \: of \: HOCN \\ 1 \: l \: contains \: 0.250 \: moles \: of \: potassium \: hydroxide \\ 0.01 \: l \: will \: contain \: (0.01 \times 0.250) \\ = 0.0025 \: moles \: of \: potassium \: hydroxide \\ KOH + HOCN → KCN + H _{2}O \\ k _{b} = \frac{[K {}^{ + } ][OH {}^{ - } ]}{[KOH]} \\ moles \: of \: OH {}^{ - } \: remaining = 0.006 - 0.0025 \\ = 0.0035 \: moles \\ total \: volume = 0.03 + 0.01 \\ = 0.04 \: litres \\ 0.04 \: l \: contains \: 0.0035 \: moles \\ 1 \: l \: will \: contain \: ( \frac{1 \times 0.0035}{0.04} ) \\ = 0.0875 \: mol \: per \: litre \\ [H {}^{ + } ] = \frac{kw}{[OH {}^{ - } ]} \\ = \frac{1 \times {10}^{ - 14} }{0.0875} \\ = 1.143 \times {10}^{ - 13} \\ pH = - log(1.143 \times {10}^{ - 13} ) \\ pH = 12.94[/tex]

Answer:

first the right answer is

Li + H2O = LiOH + H2O

then to balance it...

2 Li + 2 H2O = 2 LiOH + H2

so that we can have both side of the equation to be equal....

HOPE ITS HELPS....

Answer:

correct answer i C i think so

Answer:

Safely throw away the liquids and get new ones.

Explanation:

Not A  because

you shouldn't smell them cause it could be toxic and hurt you

Not B

you should never guess which chemical are which cause if your not correct and use the wrong one you could have a chemical reaction

Not C

you cant mix random liquids it could make a chemical reaction to exploded ore worse create a hurtful gas

Answer:

I think:

CH3COCH2CH2CH3 belongs to ketone because it has a carbonyl group C=O in its structure.

Answer:

0.034 L

Explanation:

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

Initial volume (V₁) = 0.055 L

Initial pressure (P₁) = stp = 1 atm

Initial temperature (T₁) = stp = 273 K

Final pressure (P₂) = 2.8 atm

Final temperature (T₂) = 195 °C = 195 °C + 273 = 468 K

Final volume (V₂) =?

The final volume of the gas can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

1 × 0.055 / 273 = 2.8 × V₂ / 468

0.055 / 273 = 2.8 × V₂ / 468

Cross multiply

273 × 2.8 × V₂ = 0.055 × 468

764.4 × V₂ = 25.74

Divide both side by 764.4

V₂ = 25.74 / 764.4

V₂ = 0.034 L

Therefore, the new volume of the gas is 0.034 L.

Answer:

[tex]\mathbf{FeCl_3 + AgNO_3 \to 3 AgCl+Fe(NO_3)_3}[/tex]

Explanation:

From the given question,

The chemical equation is:

[tex]\mathbf{FeCl_3 + AgNO_3 \to 3 AgCl+Fe(NO_3)_3}[/tex]

We are given that:

mass of [tex]FeCl_3[/tex] = 45 g

number of moles of [tex]FeCl_3[/tex]  = 45 g/162.2 g/mol

= 0.28 mol

mass of [tex]AgNO_3[/tex] = 25g

number of moles of [tex]AgNO_3[/tex]  = 25/169.87

= 0.147 mol

From the given equation 1 mole of [tex]FeCl_3[/tex] is required to make 3 mole of [tex]AgNO_3[/tex]  

0.28 mole of [tex]FeCl_3[/tex] = 0.28 × 3 = 0.84 mol of [tex]AgNO_3[/tex]  

Here [tex]AgNO_3[/tex]  is the limiting reagent.

Thus,

3 mole of [tex]AgNO_3[/tex]  = 3 moles of AgCl

0.147 mole of [tex]AgNO_3[/tex] = 0.147 mole of AgCl produced.

Answer:

The substances made from polyvinyls are generally brittle and hard. This property prevails because the polymer strands present in the polyvinyls form a very regular and well-organized packing composition in the solid-state. The Van der Waals associations between the strands make the substance brittle. To make the substance soft and suitable to use as a raincoat, the incorporation of small molecules is done within the molten polymer before the hardening of the plastic.

These polymer strands act as plasticizers and prevent the strands of polymer from forming a hard and brittle conformation.

The organ system has been defined as the collective working of the organs for performing a specified task. Each organ system has been assigned a major task to be performed.

The function of the following organ systems have been:

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Answer:

823.7g

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

Using the information given in this question as follows:

Q = 6,400 J

m = ?

c of soil = 0.840 J/g°C

∆T = 9.25°C

Using Q = mc∆T

m = Q ÷ c∆T

m = 6,400 ÷ (0.840 × 9.25)

m = 6400 ÷ 7.77

m = 823.7g

Answer:

There are 823.68 grams.

Explanation:

Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.

Between heat and temperature there is a relationship of direct proportionality. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body. In summary, the amount of heat Q that receives or transmits a mass m of a substance with specific heat C to raise its temperature from T1 to T2 is given by the formula:

Q= C*m* (T2- T1) = C*m* ΔT

In this case:

Replacing:

6400J= 0.840 [tex]\frac{J}{g*C}[/tex] *m* 9.25 C

Solving:

[tex]m=\frac{6400 J}{0.840 \frac{J}{g*C} *9.25 C}[/tex]

m=823.68 grams

There are 823.68 grams.

Answer:

Explanation:

Decomposition is the answer.

If a compound breaks into two or more parts then it is called decomposition.

A student enters the canteen and buys iced tea. He then goes on to add more sugar to his iced tea. In this solution, iced tea is the solvent and sugar is the solute.

Answer:

c 18.0ml

Explanation:

The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol.

Answer:

+2

Explanation:

I am first with correct answer

Answer:

the oxidation state of iron in FeC2O4 is +2

Answer:

pH = 0.90

Explanation:

The NH3 reacts with nitric acid, HNO3 as follows:

NH3 + HNO3 → NH₄⁺ + NO₃⁻

Where 1 mole of NH3 reacts per mole of HNO3.

The moles added of each reactant are:

Moles NH3:

100mL = 0.100L * (0.050mol / L) = 0.0050 moles NH3

Moles HNO3:

20mL = 0.0200L * (1.00mol / L) = 0.0200 moles HNO3

That means HNO3 is the excess reactant and, after the reaction, its moles are:

0.0200mol - 0.0050mol = 0.0150 moles HNO3

In a final volume of 120mL = 0.120L, the molar concentration of HNO3 is:

0.0150 moles HNO3 / 0.120L = 0.125M HNO3

As nitric acid is a strong acid, [HNO3] = [H+]

[H+] = 0.125M

pH = -log[H+]

Answer:

Molarity = 0.809 M

mole fraction = 0.047

Explanation:

The complete question is

Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution -

Solution for molarity:

1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

1)  

Mass of 1.09 mole of acetone

= 1.09  mol x 58.0794 g/mol = 63.306 g

Density of acetone = 0.788 g/cm3  

Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

For ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

a) Molarity:

1.09 mol / 1.347 L = 0.809 M

Mole Fraction  

a) moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

b) moles of acetone:

1.09 / (1.09 + 21.71) = 0.047

The question is incomplete, the complete question is;

And a copper electrode with 0.500 M Cu²⁺ as the second half cell

Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red= 0.337 V

The measured cell potential when the water sample was placed into the silver side of the cell was 0.0925 V.

A- What is the standard cell potential for this cell in V?

B- What is the value of the standard free energy (in kJ) for this reaction?

C- Write the balanced equation for the overall reaction in acidic solution?

D- And the measured cell potential is 0.0925, what is the concentration of chloride ions in the solution?

Answer:

See Explanation

Explanation:

a) E°cell = E°cathode - E°anode

E°cell = 0.337 V - 0.2223 V

E°cell = 0.1147 V

b) ΔG°cell = −nFE°cell

Where n=2 and F = 96500C

ΔG°cell =-(2 * 96500 *  0.1147 )

ΔG°cell =-22,137.1 J or -22.1371 KJ

c) 2Ag(s) + 2Cl⁻(aq) + Cu²⁺(aq) -----> 2AgCl(s) + Cu(s)

d) From Nernst Equation;

E= E°cell - 0.0592/n  log Q

0.0925  =  0.1147 - 0.0592/2 log 1/[0.500] [Cl⁻]^2

0.0925 - 0.1147 = - 0.0592/2 log 1/[0.500] [Cl⁻]^2

-0.0222 = -0.0296 log 1/[0.500] [Cl⁻]^2

-0.0222/-0.0296 = log 1/[0.500] [Cl⁻]^2

0.75 = log 1/[0.500] [Cl⁻]^2

Antilog (0.75) = 1/[0.500] [Cl⁻]^2

5.6234 * 0.500 =  [Cl⁻]^2

[Cl⁻] = √2.8117

[Cl⁻] = 1.68 M

A galvanic cell is a voltaic cell that produces electrical energy from the oxidation-reduction process. The standard cell potential of the cell is 0.1147 V.

The standard cell potential is a difference between the electrode potential of the cathode and the anode of the cell.

The balanced overall reaction of the cell is given as,

[tex]\rm 2Ag(s) + 2Cl^{-}(aq) + Cu^{2+}(aq) \rightarrow 2AgCl(s) + Cu(s)[/tex]

The standard cell potential is calculated as:

[tex]\begin{aligned} \rm E^{\circ}cell &= \rm E^{\circ}cathode - E^{\circ} anode\\\\&= 0.337 - 0.2223 \\\\&= 0.1147\;\rm V\end{aligned}[/tex]

Therefore, 0.1147 V is the standard cell potential for the cell.

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Answer:

1.20 × 10³ torr

Explanation:

Step 1: Given data

Step 2: Calculate the final pressure of the gas

Considering the constant volume, if we assume the gas behaves ideally, we can calculate its final pressure using Gay-Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = 822 torr × 475 K/325 K = 1.20 × 10³ torr

The pOH of the solution containing 9.8×10⁻³ M of hydrogen ion concentration, [H⁺] is 11.99

First, we shall determine the hydroxide ion concentration [OH⁻] in the solution. Details below:

[H⁺] × [OH⁻] = 10¯¹⁴

9.8×10⁻³ × [OH⁻] = 10¯¹⁴

Divide both side by 9.8×10⁻³

[OH⁻] = 10¯¹⁴ / 9.8×10⁻³

[OH⁻] = 1.02×10⁻¹² M

Finally, we shall determine the pOH of the solution. Details below:

pOH = -Log [OH⁻]

pOH = -Log 1.02×10⁻¹²

pOH = 11.99

Thus, from the above calculation, we can conclude that the pOH of the solution is 11.99

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Answer: The half-reactions represents reduction are as follows.

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in [tex]Cr_{2}O^{2-}_{7}[/tex] is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

[tex]Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}[/tex]

Similarly, oxidation state of Mn in [tex]MnO^{-}_{4}[/tex] is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

[tex]MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}[/tex]

Thus, we can conclude that half-reactions represents reduction are as follows.

Explanation:

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