All categories

2 years ago

What would cause a liquid to turn into a solid?

Chemistry

2 answers:

KIM [24]2 years ago

3 0

It’s C
Cooling a liquid til it freezes will cause a liquid to turn into a solid

mihalych1998 [28]2 years ago

3 0

C. cooling it until it freezes

You might be interested in

4. When benzene reacts with 1-chloro-2,2-dimethylpropane (neopentyl chloride) in the presence of aluminum chloride, the major pr

BARSIC [14]

Answer:

Explanation:

In this case, we have a Friedel-Craft reaction (see figure 1). Usually in this type of reaction, the benzene ring is bonded directly to the carbon that has the halide atom. But in this specific case, this is not happening.

To understand why we have to remember the reaction mechanism of this reaction. As first, step the Cl-C bond attacks the $AlCl_3$ and we form a new bond between the alkyl halide and the $AlCl_3$ , finally the C-Cl bond is broken and a carbocation is formed. In this case, we will have a primary carbocation, a very unstable species. Therefore we will have a methyl shift to obtain a tertiary carbocation.

This tertiary carbocation will react with and continues the reaction. (See figure 2)

I hope it helps!

8 0

3 years ago

How many atoms is 50 g of He?

tigry1 [53]

Atomic mass He = 4.002 u.m.a

4.002 g --------------- 6.02x10²³ atoms
50 g ------------------ ? atoms

50 x ( 6.02x10²³) / 4.002

= 3.01x10²⁵ / 4.002

= 7.52x10²⁴ atoms of He

6 0

3 years ago

Question 11 (1 point)

iren2701 [21]

Blue to red is acidic, red to blue is basic and no change is neutral.

6 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

HELP HELP HELP HELP !!!

Stella [2.4K]

A neutral atom that loses an electron becomes a positive ion. ... An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion

4 0

3 years ago

Read 2 more answers