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Answer 1

Answer:

Answer :

(a) The molecular equation will be,

$CaCO_3(aq)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$

(b) The complete ionic equation in separated aqueous solution will be,

$Ca^(2+)(aq)+CO_3^(2-)(aq)+2H^(+)(aq)+2Cl^(-)(aq)\rightarrow Ca^(2+)(aq)+2Cl^-(aq)+H_2O(l)+CO_2(g)$

(c) The net ionic equation will be,

$CO_3^(2-)(aq)+2H^(+)(aq)\rightarrow H_2O(l)+CO_2(g)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The molecular equation will be,

$CaCO_3(aq)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$

(b) The complete ionic equation in separated aqueous solution will be,

$Ca^(2+)(aq)+CO_3^(2-)(aq)+2H^(+)(aq)+2Cl^(-)(aq)\rightarrow Ca^(2+)(aq)+2Cl^-(aq)+H_2O(l)+CO_2(g)$

In this equation, $Ca^(2+)\text{ and }Cl^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

(c) The net ionic equation will be,

$CO_3^(2-)(aq)+2H^(+)(aq)\rightarrow H_2O(l)+CO_2(g)$

Related Questions

Hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water. If 20.6 g of sodium hydroxide reacts with an excess of hydrochloric acid, how many grams of sodium chloride are produced? HCl + NaOH → NaCl + H2O
It is the outermost layer of the solid portion of earth
At 1 atm, an unknown sample melts at 49.9 °C and boils at 209.5 °C. If the temperature is 0°C, what is the state of matter for the sample?
Which aqueous solution is colored?1. CuSO4(aq) 2. BaCl2(aq) 3. KCl(aq) 4. MgSO4(aq)
A 3.0 L flask containing helium at 145 mmHg is connected by a closed valve to a 2.0 L flask containing argon at 355 mmHg. When the valve is opened and the gases are allowed to mix equally in the two flasks, what is the total pressure (in mmHg) in the two connected flasks after mixing ?

3.7500*10^4+9.7100*5

Answers

Answer:

37548.55

Explanation:

3.7500*10^4+9.7100*5

3.7500*10000+9.7100*5

37500+48.55

37548.55

All the elements beyond uranium, the transuranium elements, have been prepared by bombardment and are not naturally occurring elements. The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle. Part A Complete the following equation: 10n+23892U→?+?01n+92238U→?+? Express your answer as a nuclear equation.

Answers

Answer:

¹₀n+ ²³⁸₉₂U → ²³⁹₉₃Np + ⁰₋₁e

Explanation:

Key statement;

The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle.

This is the beta particle; ⁰₋₁e

¹₀n+ ²³⁸₉₂U → Np + ⁰₋₁e

The mass number of Np;

1 + 238 = Np + 0

Np = 239

The atomic number of Np;

0 + 92 = Np + (-1)

92 + 1 = Np

Np = 93

The equation is given as;

¹₀n+ ²³⁸₉₂U → ²³⁹₉₃Np + ⁰₋₁e

Water flows over Niagara Falss at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing that the graviatational potential energy of falling water per second = mass (kg) x height (m) x gravity (9.8 m/s2), what is the power of Niagara Falls? How many 15 W LED light bulbs could it power?

Answers

Answer:

1. 176 × 10^12 W ; 78400000000

Explanation:

Given the following :

Fall rate = 2,400,000kg/s

Average height of fall = 50m

Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =

How many 15 W LED light bulbs could it power?

Recall : power = workdone / time

Workdone = gravitational potential energy

Mass of water = density * volume

Density of water = 1 * 10^3kg/m^3

Rate of fow = volume / time = 2400000

Hence,

Power = 1000 * 2,400,000 * 9.8 * 50

Power = 1176000000000

Power = 1. 176 × 10^12 W

How many 15 W LED light bulbs could it power?

1176000000000 / 15 = 78400000000

= 78400000000 15 W bulbs

How many formula units are in a 7.3 x 10-3 gram sample of lithium chloride, LiCI?

Answers

Answer: $1.0 * 10^(20)$ formula units

Explanation:

so lets start off by looking at what we have. we have $7.3 * 10^(-3)$ g of LiCl

which is called Lithium Chloride. in order to convert g to moles, we divide the g by the molar mass of Lithium Chloride.

whip out that HANDY DANDY PERIODIC TABLE man the PERIODIC TABLE WILL SAVE YOUR LIFE SOME DAY! someone will walk up to you all mean, and youll be like, "what, you tryna MUG me?" and then you whack 'em with the periodic table like BAM! GOTTEM!

okay so lets look at the periodic table and we notice that the atomic mass of Lithium is 6.941 and the atomic mass of Chlorine is 35.453. notice that in LiCl there is only one of each. so lets add 6.941 + 35.453 = 42.394 g/mol.

now look at what we were given: converting the given quantity to standard format instead of scientific format, we have 0.0073 grams of lithium chloride. we can convert this to moles by dividing it by its molar mass which is 42.394. $0.0073 / 42.394 = 1.72 * 10^(-4)$

now lets use AVOCADOS NUMBER i mean AVOGADROS NUMBER!! which is $6.02 * 10^(23)$

multiply $(1.72 * 10^(-4)) * (6.02 * 10^(23))$ and we get $10.35 * 10^(19)$ formula units.

if you want to be specific about the significant figures, notice that the given quantity in the question only has two significant figures. so we can alter our final answer to only have two sig figs. lets change it: $1.0 * 10^(20)$ formula units

HELP Which type of light is stored energy?A. Kinetic
B. Potential
C. Thermal
D. Field

Answers

Answer:

Explanation:

Answer:

Potential

Explanation:

50.0ml each of 1.0M Hcl and 1.0M Naoh at room temperature (20.0c) are mixed the temperature of the resulting Nacl solutions increase to 27.5cthe density if the resulting Nacl solutuion 1.02 g/ml
the specific heat of the resulting Nacl solutions is 4.06j/gc
calculate the heat of neutralisation of hcl and naoh in kj/mol nacl products

Answers

Answer:

62.12kJ/mol

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

You can find the released heat of the reaction and heat of neutralization (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be finded with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), thus:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that are reacting releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

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62.12kJ/mol is heat of neutralization

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