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Karo-lina-s [1.5K]
2 years ago
9

Finely ground nickel(II) hydroxide is placed in a beaker of water. It sinks to the bottom of the beaker and remains unchanged. A

n aqueous solution of hydrochloric acid (HCl) is then added to the beaker, and the Ni(OH ) 2 disappears. Which equation best describes what occurred in the beaker? A. Ni(OH ) 2 (s) + HCl(aq) → NiO(aq) + H 2(g) + HCl(aq) B. Ni(OH ) 2 (s) + 2HCl(aq) → NiC l 2 (aq) + 2 H 2 O(l) C. Ni(OH ) 2 (s) + 2 H 2 O(l) → NiC l 2 (aq) + 2 H 2 O(l) D. Ni(OH ) 2 (s) + 2 H 2 O(l) → NiCl 2(aq) + 3H 2O(l) + O 2(g)
Chemistry
1 answer:
Paraphin [41]2 years ago
5 0

Answer:C

Explanation:

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Consider the following reaction (X = Cl or Br) which statement s is are correct?
statuscvo [17]

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

5 0
2 years ago
List the number of each type of atom on the left side of the equation C5H12(g)+8O2(g)→5CO2(g)+6H2O(g) C 5 H 12 ( g ) + 8 O 2 ( g
Gala2k [10]

Answer:

Carbon=5, hydrogen=12, oxygen=16

Explanation:

Carbon=5, hydrogen=12, oxygen=16

In order to effectively count the number of atoms, we look at the equation closely and take note of the stoichiometric coefficients of each reactant as this influences the number of atoms of that element present.

For instance, oxygen is diatomic and has a stoichiometric coefficient of 8. This implies the there are sixteen atoms of oxygen altogether.

Note that the left hand side refers to the reactants side.

5 0
2 years ago
Read 2 more answers
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
Match the following names of glassware with what you would use them for.(1) Glassware used to accurately transfer small volumes.
Andrei [34K]

Answer:

(1)=(A), (2)=(B), (3)=D, (4)=C, (5)=E, (6)=F

Explanation:

(1) Glassware used to accurately transfer small volumes = (A) Graduated pipette, that is basically a glass tube with graduation of different volumes to be dispensed.

(2) Glassware used to accurately transfer a small, single volume = (B) Volumetric pipette, that is a glass tube with a central glass bulb and is used to dispense accurately an unique volume of liquid everytime.

(3) Glassware to deliver a volume not known in advance = (D) Buret (or burette), that is used to dispense slowly a volume of liquid when a titration process is needed

(4) Glassware best used when greater access to the contents is needed = (C) Beaker, that is basically a very open glass cylinder with a spout

(5) Glassware used to prevent splashing or evaporation = (E) Erlenmeyer flask, that has a small open at the top and is useful when the liquid needs to be swirled as, for example, during a titration.

(6) Glassware used to make accurate solutions = (F) Volumetric flask, that has a long slim neck that provides a higher accuracy when a exact volume of liquid needs to be used for preparation of a solution.

8 0
2 years ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0
2 years ago
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