All categories

2 years ago

A student melts 50.0 grams of ice using 16650 J of energy. What is the heat of fusion of ice?

Chemistry

1 answer:

Arlecino [84]2 years ago

7 0

heat of fusion = 16650 J / 50 g = 333 J/g

Thus, the heat of fusion of the ice is 333 J/g.

You might be interested in

During a reaction in an aqueous solution, the concentration of bactants

STALIN [3.7K]

Answer:

my define it will be turst me is c

5 0

3 years ago

Which statement about chemical reactions is not true?

igomit [66]

Answer:

1: New atoms are formed as products

Explanation:

matter (atoms) cannot be created or destroyed

unless you are God or Cinderella's fairy Godmother or something

7 0

2 years ago

What is the mass of a titanium alloy that has a heat of fusion of 422.5 joules/gram and requires 2,960 joules of energy to melt

romanna [79]

Answer:- Mass of the titanium alloy is 7.01 g, choice C is correct.

Solution:- The heat of fusion is given as 422.5 joules per gram and it also says that 2960 joules of heat is required to melt the metal completely.

The suggested equation is, $Q=mH_f$

where Q is the heat energy, m is the mass and Hf is the heat of fusion.

Since, we are asked to calculate the mass, the equation could be written as:

$m=\frac{Q}{H_f}$

Let's plug in the values in it:

$m=\frac{2960J}{\frac{422.5J}{g}}$

m = 7.01 g

So, the mass of the titanium alloy is 7.01 g, choice C is correct.

7 0

2 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

What is the precipitate for CuSO4+NaOH<br> (Balancing Equations)

jeka57 [31]

Answer:

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

Explanation:

5 0

3 years ago