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Gelneren [198K]
3 years ago
6

Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero d

egrees Celsius, melts at zero degrees Celsius, and boils at 100 degrees Celsius. Remember that you need to take into account three changes: melting ice, heating water, and vaporizing the water.)
Chemistry
1 answer:
Yuki888 [10]3 years ago
7 0
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g 
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK 
</span>
Now, energy required for melting of ICE = <span>  334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC =  4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span>  2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2)  + (3)
                                                                        = 1753.5 + 2195.18 + 11849.25
                                                                        = </span><span>15797.93 J 
</span><span>                                                                        = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
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When the following equation is balanced using the smallest possible integers, what is the coefficent of oxygen gas?
elena55 [62]

By balancing a chemical reaction, you must first balance the atom that is in smaller amount in the chemical species that you have in the reaction. In the combustion reaction presented in the question that atom is carbon (C). Therefore, the first thing to do is multiply the CO2 by 7, since those are the carbon atoms that are in the C7H16O. The reaction would be like this,

C7H16O + O2  →  7CO2 + H2O

You can make a table of the amount of atoms of C, H and O that you have in the reagents and products after you put a coefficient 7 in front of the CO2, in the following way,

R P

C 7 7

H 16 2

O 3 15

Then you balance the hydrogens. It is better that you leave the last oxygens since there is an oxygen molecule alone, so when adding a coefficient to balance it, the quantities of the rest of the atoms in the equation would not be altered.

To balance the hydrogens you add an 8 in front of the H2O molecule in the reagents, since there are 16 hydrogens in the molecule C7H16O and

8H x 2H = 16H.

The reaction and the table would be like this,

C7H16O + O2  →  7CO2 + 8H2O

R P

C 7 7

H 16 16

O 3 22

Finally, you balance the atoms of O. To do this, you add a coefficient of 21/2 to the O2 molecule, this is because (21/2) x2 = 21 and 21 O coming from the O2 molecule plus an O coming from the C7H16O molecule gives a total of 22 O, which are equal to the amount of O we have in the products.

C7H16O + (21/2)O2  →  7CO2 + 8H2O

To bring the coefficients of the reactants and the products to whole numbers multiply all the coefficients by 2. Then the reaction is like,

2 C7H16O + 21O2  →  14CO2 + 16H2O

As you can see when the given reaction is balanced using the smallest possible integers, the coefficent of oxygen gas is 21

6 0
3 years ago
Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:
tekilochka [14]

Answer:

See explanation

Explanation:

First, let's write the balanced equation again:

2 NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)

Now, we know that the total pressure was 7.76 atm. This total pressure, is the sum of the pressure of water and CO2 like this:

Ptotal = Pwat + PCO2 (1)

This is the dalton's law for partial pressures.

The pressure can be also be relationed with the moles

Ratio of mole = Ratio of pressure

so, taking this in consideration we can say the following:

Pwater/PCO2 = moles water / moles CO2

As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

With this pressure, and using the ideal gas equation, we can know the moles of water:

PV = nRT

n = PV/RT     using R = 0.082 L atm / K mol

n = 3.88 * 5 / 0.082 * (160+273)

n = 0.546 moles of water

b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)    MMCO2 = 84 g/mol

the moles of NaHCO3 initially:

n = 100 / 84

n = 1.19 moles

so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:

remaining moles = 1.19 - 0.546 = 0.644 moles

therefore the mass remaining:

mCO2 = 0.644 * 84

mCO2 = 54.096 g

c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:

Kp = Pwater * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.

3 0
3 years ago
Malonic acid, h2c3h2o4, is a diprotic acid with ka1= 1.5 x 10-3and ka2= 2.0 x 10-6. what is the concentration of the malonate an
Bess [88]

In a 0.15 m solution of malonic acid, the amount of the malonate anion, c3h2o42-, is 1.68*10^{-4} M.

Explanation:

For the first equlibria;

initial concentration of H_{2}C_{3}H_{2}OH=0.15

initial concentration of HC_{3}H_{2}O_{4}^{-} =0

initial concentration of H^{+} =0

Let concentration change of H_{2}C_{3}H_{2}OH=-x

Let concentration change of HC_{3}H_{2}O_{4}^{-} =+x

Let concentration change of H^{+} =+x

Let concentration in equilibrium state of  H_{2}C_{3}H_{2}OH=0.15-x

Let concentration in equilibrium state ofHC_{3}H_{2}O_{4}^{-} =x

Let concentration in equilibrium state ofH^{+} =x

Therefore, Ka_{1}=\frac{x*x}{0.15-x}=\frac{1.5*10^{-3}}{1} =\frac{x^{2}}{0.15-x}

x^{2}+0.0015x-0.000225=0

Which is a quadratic equation in x, solving it gives;

x=0.01427

Therefore, HC_{3}H_{2}O_{4}^{-} =x= 0.01427M

Now for the second equlibria;

initial concentration of HC_{3}H_{2}O_{4}^{-}=0.01427

initial concentration of C_{3}H_{2}O_{4}^{2-} =0

initial concentration of H^{+} =0

Let concentration change of HC_{3}H_{2}O_{4}^{-}=-x

Let concentration change of  C_{3}H_{2}O_{4}^{2-} =+x

Let concentration change of H^{+} =+x

Let concentration in equilibrium state of  HC_{3}H_{2}O_{4}^{-}=0.01427-x

Let concentration in equilibrium state ofC_{3}H_{2}O_{4}^{2-} =x

Let concentration in equilibrium state ofH^{+} =x

Therefore, Ka_{2}=\frac{x*x}{0.01427-x}=\frac{2.0*10^{-6}}{1} =\frac{x^{2}}{0.01427-x}

2.854*10^{-8}-(2.0*10^{-6})x=x^{2}

Therefore, C_{3}H_{2}O_{4}^{2-} =x= 1.68*10^{-4}M

<h3>What is a malonic acid?</h3>

The chemical formula of malonic acid is CH2(COOH)2. Malonates include the ionized form of malonic acid as well as its esters and salts. For instance, the diethyl ester of malonic acid is called diethyl malonate.

Polyester and polymers both employ malonic acid as a precursor. Both the scent business and the flavoring sector utilize it. It is employed to regulate acidity. Malonic acid contributes to the production of fatty acids in people. Malonic acid has been found in a number of foods outside of the human body, including red beets, corn, scarlet beans, common beets, and cow milks, although its presence has not been defined. The group of organic substances known as dicarboxylic acids and derivatives includes malonic acid, also known as malonate or H2MALO.

To know more about malonic acid, visit:

brainly.com/question/13943912?referrer=searchResults

#SPJ4

4 0
1 year ago
Explain step by step, please!
konstantin123 [22]

Answer:

10. Stacey

11. 7.86 g/cm^3 (3dp)

OR 786 kg/m^3 (SI units)

Explanation:

10. The correct exact answer for 10 is 26.169 (6.71x3.9). While Sam was correct in the precision of the number (2dp just like the given measurements), he was incorrect in the number itself. While Stacy's number was not as precise as Sam's (0dp), it is correctly rounded to the nearest whole number, therefore Stacy is correct.

11. The formula for density is mass/volume. The mass here is 264g and the volume is 33.6ml, therefore the density is 7.86 (3 sig figs because given values had 3) g/cm^3, because the mass was given in grams and 1ml = 1cm^3. Converted to SI units it is 786 kg/m^3.

Hope this helped!

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3 years ago
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tensa zangetsu [6.8K]

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HOPE THIS HELPS

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