Ask question

Ask question

All categories

3 years ago

10

A typical phase change diagram is shown below. Which of the numbered arrows represent changes where the kinetic energy of the pa

rticles increases

1 answer:

NikAS [45]3 years ago

7 0

Do you have a diagram?

You might be interested in

Which answer below correctly identifies the type of change and the explanation when magnesium comes into contact with hydrochlor

Oliga [24]

Answer:

A chemical change because a temperature change occurred, the solid disappeared and a gas was produces

Explanation:

Magnesium reacts with hydrochloric acid releasing energy, and leading to the formation of magnesium chloride and hydrogen gas. This is represented by the equation below:

Mg₍s₎ + 2HCl₍aq)⇒ MgCl₂₍aq₎ + H₂₍g₎

5 0

3 years ago

Read 2 more answers

An object becomes electrically charged when:

timurjin [86]

Answer:

An object becomes electrically charged when <em>electrons</em><em> </em><em>are</em><em> </em><em>transferred</em><em> </em><em>to</em><em> </em><em>it</em>

Explanation:

<em>PLEASE</em><em> </em><em>DO MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>

7 0

3 years ago

Anyone know ?? I need help with this

nevsk [136]

Answer:

A

Explanation:

Because the amount of electrons in the outer shell can be a maximum of 8

Hope this helps

6 0

2 years ago

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

5 0

2 years ago

If you change the 2 in front of 2O2 to a 3, what will be the change in the results on the right side of the equation? (1 point)

Simora [160]

Answer:

There is an extra O2 molecule left over

Explanation:

3 0

2 years ago