A(n) ____________________ has a rectangular shape and contains a cell wall, chloroplasts, and one large vacuole.

Use the particle theory to explain,liquid to soild process

hjlf

The liquid to solid process using the particle theory is as below

The process that involve change of liquid to solid is known as Freezing
It involve change from a high energy state to lower energy

The constant temperature at which a liquid change to solid by giving out heat energy is called freezing point of the liquid
when liquid are cooled the thermal energy of particles decrease.
The cohesive forces between the particles strengthen to such extent that particles can have relative motion with each other and they occupy the fixed position, thus liquid is converted to solid

4 0

3 years ago

Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 ×

xz_007 [3.2K]

Answer:

K, the rate constant = 9.73 × 10^(-1)/s

Explanation:

r = K × [A]^x × [B]^y

r = Rate = 1.07 × 10^(-1)/s

K = Rate constant

A and B = Concentration in mol/dm^-3

A = 0.44M

B = 0.11M

x = Order of reaction with respect to A = 0

y = Order of reaction with respect to B = 1

Solving, we get

r/([A]^x × [B]^y) = K

K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727

K = 0.9727

7 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0

3 years ago

I NEED HELP ASAP! WILL MARK AS BRAINLIEST

LiRa [457]

Answer:

Sry i am unable to see the attached picture but i hope this helps

Explanation:

There are a couple of ways to prepare a buffer solution of a specific pH. In the first method, prepare a solution with an acid and its conjugate base by dissolving the acid form of the buffer in about 60% of the volume of water required to obtain the final solution volume

5 0

3 years ago

Read 2 more answers

Which of the following statements is true?

liubo4ka [24]

Answer:

a is the anwser

Explanation:

4 0

3 years ago