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3 years ago

What happens when a solid becomes a liquid?

Chemistry

2 answers:

tatiyna3 years ago

8 0

Answer:

B. It changes shape but not volume.

APEX

Mnenie [13.5K]3 years ago

5 0

All its doing is changing its physical state so it chemical state not being changed means its still the same thing it was as a solid, just now its turned into a liquid.

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Na2O + HCl -> NaCl + H2O

Charra [1.4K]

Na2O + HCl→ NaCl +H2O

Does the equation above satisfy the law of mass conservation of matter.

No, since there are two sodium atoms as reactants, but only one sodium atom as a product.

Explanation

According to the law of mass conservation the number of atoms in the reactant side must be equal to the number of atoms in the product side.

For this reason the equation above does not satisfy the law of mass conservation since the number of sodium atoms are not equal in both side.

6 0

3 years ago

What sign shoes dissolving

almond37 [142]

Answer: can u explain ur question please?

Explanation:

8 0

3 years ago

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Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)

riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

$2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)$

The expression used for entropy change of reaction $(\Delta S^o)$ is:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]$

where,

$\Delta S^o$ = entropy change of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

$\Delta S_f^0_{(Br_2)}$ = 245.463 J/mol.K

$\Delta S_f^0_{(F_2)}$ = 202.78 J/mol.K

$\Delta S_f^0_{(BrF_3)}$ = 292.53 J/mol.K

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]$

$\Delta S^o=268.74J/K$

Now we have to calculate the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of $BrF_3$ has entropy change = 268.74 J/K

So, 1.62 moles of $BrF_3$ has entropy change = $\frac{1.62}{2}\times 268.74=217.68J/K$

Therefore, the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

3 0

4 years ago

a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

1223.38 mmHg must be the pressure of the nitrogen gas.

5 0

3 years ago

✓7 Is it rational or irrational

Elden [556K]

Answer:

Irrational

Explanation:

5 0

4 years ago

Read 2 more answers