All categories

3 years ago

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.

Chemistry

1 answer:

Goshia [24]3 years ago

5 0

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

You might be interested in

How many significant figures are in 0.0233?

viva [34]

Answer:

there are 3 significant figures

Explanation:

do not count the first 2 zeros. only the nimbers after the zero

7 0

3 years ago

How many grams of CH4 will be in a 500ml contain at STP?

ArbitrLikvidat [17]

Answer:

Mass = 0.32 g

Explanation:

Given data:

Mass of CH₄ = ?

Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)

Temperature = 273 K

Pressure = 1 atm

Solution:

Volume of CH₄:

500 mL (500 mL× 1L/1000 mL= 0.5 L)

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

By putting values,

1 atm× 0.5 L = n×0.0821 atm.L/ mol.K × 273 K

0.5 atm.L = n×22.4 atm.L/ mol

n = 0.5 atm.L / 22.4 atm.L/ mol

n = 0.02 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.02 mol × 16 g/mol

Mass = 0.32 g

7 0

3 years ago

Why doesnt iron burst into flames when it rusts

Katarina [22]

Rust does not have the properties to catch onto flames. However, if you light it on fire, then it would probably catch in flames but not instantly and will not continue to burn unless you have soaked it in oil or flammable object or substance. :) Hope this helps!

3 0

3 years ago

Counseling psychologist generally deal with more serious problems then do clinical psychologists

ElenaW [278]

This can be seen in many ways depending on the situation at hand. Also the concern the counselor has for the problem is a key factor.

8 0

3 years ago

If 6.00 g of CaCl2 • 2 H2O and 5.50 g of Na2CO3 are allowed to react in aqueous solution, what mass of CaCO3 will be produced? P

Andre45 [30]

Answer:

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of CaCl₂ .2H₂O × 100 g CaCO₃ / 1 mole CaCO₃ = 4 g

5.50 g Na₂CO₃ /1 × 1 Na₂CO₃ / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃ × 100 g CaCO₃ / 1 mole CaCO₃ = 5 g

Explanation:

Given data:

Mass of CaCl₂.2H₂O = 6.00 g

Mass of Na₂CO₃ = 5.50 g

Mass of CaCO₃ produced = ?

Solution:

Number of moles of CaCl₂.2H₂O.

Number of moles = mass/ molar mass

Number of moles = 6.00 g/ 147 g/ mol

Number of moles = 0.04 mol

Number of moles of Na₂CO₃:

Number of moles = mass/ molar mass

Number of moles = 5.50 g/ 106 g/ mol

Number of moles = 0.05 mol

Chemical equation:

CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl

Now we will compare the moles of CaCO₃ with Na₂CO₃ and CaCl₂ through balanced chemical equation .

CaCl₂ : CaCO₃

1 : 1

0.04 : 0.04

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.04 mol× 100 g/mol

Mass = 4 g

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of CaCl₂ .2H₂O × 100 g CaCO₃ / 1 mole CaCO₃ = 4 g

Na₂CO₃ : CaCO₃

1 : 1

0.05 : 0.05

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.05 mol× 100 g/mol

Mass = 5 g

5.50 g Na₂CO₃ /1 × 1 Na₂CO₃ / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃ × 100 g CaCO₃ / 1 mole CaCO₃ = 5 g

5 0

3 years ago