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dalvyx [7]
2 years ago
10

Determine the mass density, specific volume and specific weight of a liquid whose

Chemistry
1 answer:
irakobra [83]2 years ago
5 0

the volume is 63 i hope it is

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A sample of water is heated from 60.0 °C to 75.0°C by the addition of 140 j of
kupik [55]

Mass of the water : 2.23 g

<h3>Furter explanation</h3>

Heat

Q = m.c.Δt

m= mass, g

c = heat capacity, for water : 4.18 J/g° C.

ΔT = temperature

Q= 140 J

Δt = 75 - 60 = 15

mass of the water :

\tt m=\dfrac{Q}{c.\Delta T}=\dfrac{140}{4.18\times 15}=2.23~g

5 0
3 years ago
Name three of the most sensitive parts of your body (Appropriate parts. Ex: finger, wrist, shoulder, etc.)
yKpoI14uk [10]
Inner thigh, eyes, brain
3 0
3 years ago
Consider the following intermediate reactions.
Alja [10]

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

4 0
4 years ago
Read 2 more answers
Explain how the bonding model for sodium metal would differ from the bonding model for sodium chloride, NaCi
Elina [12.6K]
Sodium metal is quite reactive; sodium ions (as in NaCl) are quite unreactive. Cl^- ions are not reactive; they are stingily attracted to positive ions such as Na^+with which they form ionic bonds.
8 0
3 years ago
The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer
sergey [27]

Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

pH=pKa+log(\frac{salt}{acid} )

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

8 0
3 years ago
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