Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.
P4(s) + NaOH(aq) + H2O(l) -> Ph3(g) + Na2HPO3(aq)

Answers

Answer 1

Answer:

I used a,b c, d in the equation as substituted coefficients to find the unknown for each element of P, Na, O, H, and I got

P4(s) + 4NaOH(aq) + 2H20(l)---->2Ph3 +2Na2HPO3(aq).

which I think should be the answer.


Related Questions

When should a line graph be used

Answers

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.


Which redox reaction would most likely occur if zinc and copper metal were
added to a solution that contained zinc and copper ions?
Click for a reduction potential chart
A. Cu + Zn → Cu2+ + Zn2+
B. Cu + Zn2+
Cu2+ + Zn
C. Cu2+ + Zn → Cu + Zn2+
D. Cu2+ + Zn2+ → Cu + Zn

Answers

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.

What is Redox Reaction?

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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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 metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?

Answers

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.

What is a calorimeter?

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,

Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.

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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Briefly workout the relationship between these constants:
[tex]{ \bf{K _{sp} \: and \: K _{c} }}[/tex]
In consideration of the decopmposition of hydrogen iodide.
[tex]{ \sf{2HI _{(g)} →H _{2(g)} +I _{2(g)} }}[/tex]
[tex]{ \tt{any \: help \: is \: appreciated}}[/tex]

Answers

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.


What is the main reason for using a data table to collect data?
A. To interpret the possible meaning of the data
B. To find the possible errors that were made in recording the data
C. To organize the information so that it is easier to understand
O
D. To make an experimental journal more attractive

Answers

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

A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.

Answers

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]

What separates the inner planets from the outer planets?

a. Main asteroid belt
b. Main comet belt
c. Kuiper belt
d. Outer planet belt
please help this is for SCIENCE test I need help

Answers

Answer:

main asteroid belt separates the inner planets from the outer planets

Calculating the expected pH of the buffer solution: Given that the pKa for Acetic Acid is 4.77, calculate the expected pH of the buffer solutions using the Henderson-Hasselbalch equation and the concentrations of Acetic Acid and Acetate added to the 250 ml Erlenmeyer flask: pH

Answers

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]

A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?

Answers

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

An unknown compound has the following chemical formula:
Co(OH),
where x stands for a whole number.
Measurements also show that a certain sample of the unknown compound contains 5.1 mol of oxygen and 2.59 mol of cobalt.
Write the complete chemical formula for the unknown compound.

Answers

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.

Consider the molecule PF5.
Indicate how many lone pairs you would find on the central atom:
Indicate how many total bonds are connected to the central atom (count single bonds as 1 bond, double bonds as 2 bonds, and triple bonds as 3 bonds):

Answers

Explanation:

here's the answer to your question

2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate

Answers

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.

consider the following thermochemical reaction for kerosene
2C12H26+37O2=24CO2+15026kj.
a. when 21.3g of CO2 are made, how much heat is released?
b. if 500.00kj of heat are released by thye reaction, how many grams of C12H26 have been consumed.?
c. if this reactionwere being used to generate heat, how many grams of C12H26 would have to be reacted to generate enough heat to raise the temperature of 750g of liquid water from 10 degrees celcius to 90 degrees celcius

Answers

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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Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

Answers

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%

How many grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride?

Answers

Answer:

it is 11.55 and ik because I just had that question

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:

[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]

The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:

[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

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19. Which type of chemical process is used to remove salt from ocean water?
O A. Alkylation
O B. Doping
O C. Dehydrogenation
D. Desalination

Answers

Answer:

D

Explanation:

Desalination

Removing salt from sea water is known as desalination

A tree is an example
of a vascular plant that
is
because it
has deep roots.
A. tall
B. tiny
C. small
Dyshort

Answers

The answer is A it’s tall

explain hydrogen dioxide​

Answers

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

Determine the number of water molecules in 0.2830g Na. ​

Answers

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

You need to make an aqueous solution of 0.121 M magnesium acetate for an experiment in lab, using a 250 mL volumetric flask. How much solid magnesium acetate should you add

Answers

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

In order to complete the reaction of hexyl magnesium bromide with acetone, what next step needs to be done.
a. Fractional Distillation.
b. Vacuum filtration.
c. Aqueous workup.
d. Crystallization.

Answers

Answer:

Aqueous workup.

Explanation:

The reaction of hexyl magnesium bromide with acetone yields a tertiary alcohol. There is an organic phase and an aqueous phase.

Aqueous workup is the process of recovering the pure tertiary alcohol from the organic phase of the system.

Hence, in order to complete the reaction of hexyl magnesium bromide with acetone, aqueous workup is required.

The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g.oC. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained.
Type of metal used:
Trial 1 Trial 2 Trial 3
Mass of metal, g 2.746 g 2.750 g 2.900 g
Mass of water, g 15.200 g 15.206 g 15.201 g
Initial Temp. of Water, oC 24.7 oC 24.6 oC 24.5 oC
Initial Temp. of Metal, oC 72.1 oC 72.2 oC 71.9 oC
Final Temp of Water & Metal,oC 26.3 oC 26.2 oC 24.7 oC
ΔT for water, oC ______ ______ ______
ΔT for metal, oC ______ ______ ______
Specific heat capacity of metal, J/g.oC ______ ______ ______
Average specific heat capacity, J/g .oC ______ (use two significant figures due to ΔT of water)

Answers

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

A student was given a solid containing a mixture of nitrate salts. The sample completely dissolved in water, and upon addition of dilute HCl , no precipitate formed. The pH was lowered to about 1 and H2S was bubbled through the solution. No precipitate formed. The pH was adjusted to 8 and H2S was again bubbled in. This time, a precipitate formed. Which compounds might have been present in the unknown?
a. Ca(NO3)2
b. AgNO3
c. Fe(NO3)3
d. Cr(NO3)3
e. Cu(NO3)2
f. KNO3
g. Bi(NO3)2

Answers

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.

Why does the temperature stop rising while ice melts into water?
A. The temperature does not stop rising.
B. The electrons are increasing in energy levels.
C. Because no more heat is being added to the system.
D. The energy is being absorbed to separate the particles.

Answers

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.

What is 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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Use the reaction: 2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq) What volume (mL) of 0.568 M AgNO3(aq) is needed to form 0.21 g of Ag2SO4(s)

Answers

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.

Given the following balanced equation:
3Cu(s) + 8HNO3(aq) = 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
Determine the mass of copper (II) nitrate that would be formed from the complete reaction
of 35.5g of copper with an excess of nitric acid.

Answers

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.

Which species is the conjugate base of H2SO3

Answers

Explanation:

As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.

The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.

In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.

Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ

How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?

Answers

Answer:

endet nach selam nw

4gh7

How many moles of p are needed to react with 30.1 moles of O2 SHOW the math below.

Answers

Answer:

information is missing

Explanation

reaction is needed to solve the problem

What is the molecule shown below?
A. Pentane
B. Trimethylethane
C. 2,2-dimethylpropane
D. 3-dipropane


Q2​

Answers

Answer:

C

Explanation:

if we were to followw the IUPAC

Other Questions
PLEASE ANSWER ILL MARK !! Prove that the value of the expression: (3^5 - 3^4)(3^3 + 3^2) is divisible by 24. Answer format: The value of the expression is 24 __ Hi Friends!Please help me out with this question given below: Why is erosion considered a slow process? Which of the following composition of transformations would create animage that is not congruent to its original image?A rotation of 45 followed by a reflection across the x-axisA translation 2 units to the left followed by dilation by 1/2A reflection across the y-axis followed by a rotation of 60A rotation of 135 followed by a translation of 4 units to the right answeredThe presence of AIDS in a community often increase their level of poverty. what is the main reason for this ? a. superstitionb. Inability for affected people to workc. poor drinking waterd. pooor public health servicesHELP ASAP SOMEONE HELP ME !!! The side length of a smaller square is one-third the side length of a larger square.eethe following statements describes the area of the smaller square?F The area of the smaller square is I the area of the larger square.27G The area of the smaller square is 172 the area of the larger square.H The area of the smaller square is the area of the larger square.1 The area of the smaller square isthe area of the larger square,31 / 2 SOMEONE HELP ME !!! Select one of the following poems, and in a response of approximately 100 words, Identity and describe two differentkinds of figurative language used in the poem."Caged Bird" by Maya Angelou"Cloud" by Sandra Cisneros"Women" by Allce Walker"The Journey" by Mary Ollver Use labyrinthine in a sentence Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is ______ oC and the boiling point of toluene is ______ oC. Therefore, the liquid collected first should be ______Please put an answer in each box. 3. Tell whether each statement is true or false Explain how you know a) LCM (7, 18) - LCM (14.18) b) LCM (5,8) - LCM (10,8) c) The GCF of any two prime numbers is 1 and the number itself. After the increase the price of a rail season ticket from brighton to london was 2828.80 work out the price before this increase ] Widget manufacturing Company began operations on January 1. All sales are on credit. Widget has sales budgeted as $160,000 for January and $290,000 for February. Accounts Receivable collections are expected to be 60% in the month of sale, 30% the next month, and 10% in the third month. Use this information to determine the dollar value of February Expected Cash Collections from Customers. Enter as a whole number (no cents). g State if is they are functions or not by filling in the blanks yes or no. What is the smallest 6-digit palindrome (a number that reads the same forward andbackward) divisible by 99? The following condensed balance sheet is for the partnership of Hardwick, Saunders, and Ferris, who share profits and losses in the ratio of 4:3:3, respectively:Cash $83,000 Accounts payable $208,000Other assets 765,000 Ferris, loan 44,000Hardwick, loan 34,000 Hardwick, capital 280,000Saunders, capital 180,000Ferris, capital 170,000Total assets $882,000 Total liabilities and capital $882,000The partners decide to liquidate the partnership. Forty percent of the other assets are sold for $240,000. Prepare a proposed schedule of liquidation at this point in time. Suppose that the probability distribution for birth weights is normal with a mean of 120 ounces and a standard deviation of 20 ounces. The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is [ Select ] 68%. The probability that a randomly selected infant has a birth weight between 110 and 130 is [ Select ] 68%. the Battle of Bunker Hill took place shortly after what event If we plot John's opportunity cost per window on the vertical axis and the number of windows cleaned each day on the horizontal axis, we will have John's ______ curve for window-cleaning services. The lifetime of a particular brand of A batteries follows a normal distribution. The mean lifetime of particular brand of A batteries is 1000 hours, with a standard deviation of 100 hours. 1. What percentage of batteries last more than 1100 hours?a. 2.5%.b. 5%.c. 16%.d. 32%.2. What is the probability that a randomly selected battery lasts more than 875 hours?a. 0.3749.b. 0.8944.c. 0.8716.d. 0.1056.3. What is the probability that a randomly selected battery lasts between 1150 and 1250 hours?a. 0.2417.b. 0.4332.c. 0.9915.d. 0.3085. If calcium carbonate (Cacos) decomposes. What would the product of the reaction be?