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3 years ago

Question 1 of 7

Chemistry

2 answers:

Nata [24]3 years ago

8 0

Answer:

Explanation:

nlexa [21]3 years ago

3 0

Answer:

Explanation:B ice melting is a phiyscal change and the mass of the substance remains the same

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How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe

polet [3.4K]

First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L

4 0

4 years ago

If one benzene molecule is placed in water, the total entropy of the water + benzene system:___________

valkas [14]

Answer:

d) increases

Explanation:

Benzene is an aromatic hydrocarbon which is obtained from the destructive distillation of coal. It is a colourless volatile liquid with a sweet smell. It boils ar 80° C (353 K) and freezes at 5°C (2278 K). It is insoluble in water but mixes in all proportions with ethanol, ethoxyethane and methylbenzene. The reason which benzene is insoluble in water is that benzene is a non-polar compound and water is polar, meanwhile only "like dissolves like". So, when the benzene molecule is placed in water. There will be distortion and disturbance between the benzene molecule and the water. Thus, the particle of each molecule will be distant from each other. This state results to change in the entropy of the system as the entropy of the system increases.

3 0

3 years ago

A calorimeter made from a can is used to measure the calories in a peanut. During the experiment, the

kari74 [83]

Answer:

The heat generated per gram of sample in calories is 0.87 g

Explanation:

Given that,

A calorimeter made from a can is used to measure the calories in a peanut.

During the experiment, the measured calories were 35% of the true/accepted value given by references

Suppose, the mass of peanut is 0.4 g and find the

We need to calculate the heat generated per gram of sample in calories

Using formula of heat generated

$q=\dfrac{q_{penut\ calorimeter}}{m_{peanut}}$

Put the value into the formula

$q=\dfrac{0.35}{0.4}$

$q=0.87\ g$

Hence, The heat generated per gram of sample in calories is 0.87 g

3 0

3 years ago

1. How many miles are in 1,000,000 centimeters

Julli [10]

There are 6.21371192 miles in 1mil cm

5 0

3 years ago

Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G

dangina [55]

Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of water = 45 g

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of liquid water = $1.84J/g^oC$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]$

$\Delta H=139287.75J=139.28775KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

3 0

3 years ago