All categories

3 years ago

BRAINLIEST TO THE BEST ANSWER Whats it called when Solid Ice changes into liquid water?

Chemistry

1 answer:

Karo-lina-s [1.5K]3 years ago

8 0

Answer: it’s called a phase change

Explanation: the Solid ice was melted and went through a phase change to turn into liquid matter.

You might be interested in

You have a 5-liter container with 2.00 x 1023 molecules of ammonia gas (NH3) at STP. How many molecules of argon gas (Ar) are in

svet-max [94.6K]

Answer:

C. 2.00 x 1023 molecules

Explanation:

Given , the pressure , volume and temperature are same in both cases.

From the ideal gas equation,

P V = n R T

where n= no. of moles = given weight ÷ atomic weight

And 6.023 x 10^23 molecules of ammonia(NH3), weigh 17g ( 14 + 3(1 ))

So we have

P V = ((2.00 x 10^23)/(6.023 x 10^23)) x R T

Similarly, say n molecules of argon gas (Ar) are in an identical container.

So the gas equation would be

P V = ( n / 6.023 x 10^23 ) x R T

Because, atomic weight of Argon is beared by 6.023 x 10^23 molecules.

dividing these two equations, we get

n = 2.00 x 10^23 molecules

8 0

3 years ago

Calculate Δ H ∘ in kilojoules for the reaction of ammonia (Δ H ∘ f =−46.1 kJ/mol) with O 2 to yield nitric oxide NO (Δ H ∘ f =91

Vitek1552 [10]

Ummmmmmm......... im in 6th grade and i do not understand 1 bit of this, but i tried, sorry! :)

7 0

3 years ago

Which of the following gas samples would have the largest number of representative particles at STP? A. 0.520 L SO3 B. 0.10 L Xe

egoroff_w [7]

Answer: The correct option is D.

Explanation: To calculate the larges number of partivles of gas, we will use Avogadro's Law which says that volume of the gas is directly related to the number of moles of gas at constant pressure and temperature.

Mathematically,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ....(1)

The gas which has the largest number of moles, will have the largest number of particles.

We are provided to use STP conditions, which says that

1 mole of a gas occupies 22.4 L of volume.

Initial conditions:

$n_1=1mole$

$V_1=22.4L$

A. 0.520 L of $SO_3$

$V_2=0.520L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{0.520}{n_2}$

$n_2=0.0232moles$

B. 0.10 L of Xe

$V_2=0.10L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{0.10}{n_2}$

$n_2=0.0044moles$

C. 7.0 L of $N_2$

$V_2=7.0L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{7.0}{n_2}$

$n_2=0.3125moles$

D. 12.0 L of Ne

$V_2=12.0L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{12.0}{n_2}$

$n_2=0.5357moles$

Hence, from the above calculations, we see that option D has the largest amount of moles and hence, will have the largest amount of particles.

6 0

4 years ago

Read 2 more answers

Please help me guys to solve this problem!

steposvetlana [31]

(a) distance measured with metre rule or tape rule

time measured with stopwatch/ stop clock/ timer

Average speed = total distance / time

(b) (i) decrease in speed

(ii) Change in speed = a * t

4.5 m/s

hope this helps......

6 0

3 years ago

Why pie bond are not perticipate in hybridaization

DedPeter [7]

The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.

3 0

3 years ago