Ok so i haven't been on this in a long time ig and so i don't know i just need help on this so i can play my fav video game oof

How many pounds of ice are required to absorb 4900 kJ of heat as the ice melts? The heat of fusion of water is 0.334 kJ/g.

Goshia [24]

Answer:

m = 32.34 pounds of ice.

Explanation:

In this case we need to use the following expression of heat:

q = m * ΔHf (1)

Where:

q: heat absorbed in J or kJ

m: mass of the compound in g

ΔHf: heat of fusion of the water in kJ/g

We are asked to look for the mass of ice in pounds, so after we get the grams, we need to convert the grams to pounds, using the following conversion:

1 pound --------> 453.59 g (2)

So, we have the heat and heat of fusion, from (1) let's solve for the mass, and then, using (2) the conversion to pounds:

q = m * ΔHf

m = q / ΔHf

m = 4900 / 0.334 = 14,670.66 g of ice

Now, the conversion to pounds:

m = 14,670.66 g * 1 pound/453.59 g

<h2>m = 32.34 pounds of ice.</h2>

Hope this helps

5 0

3 years ago

What can Be done to reduce the negative impact of pollution

Snowcat [4.5K]

Get a majority of the world to go green

3 0

3 years ago

How many molecules are in 5.60 L of oxygen gas

topjm [15]

5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.

3 0

3 years ago

Read 2 more answers

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$