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seropon [69]
3 years ago
10

Water can dissolve other substances. Which factor best contributes to this property of water?

Chemistry
2 answers:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

D i just did the quiz

Explanation:

natali 33 [55]3 years ago
3 0

Answer: -

Water can dissolve other substances. Which factor best contributes to this property of water?

d. Water particles are not fixed in place.

Explanation: -

When a substance like say common salt sodium chloride NaCl is dissolved in water, the water molecules surround the Na⁺ and Cl⁻ which result from dissolving of NaCl in water.

The water molecules by surrounding the Na⁺ and Cl⁻ help in dissolving NaCl.

Thus in order to surround them,the water molecules should not be fixed in their place and be free to move about.

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Calculate the specific heat capacity of a piece of ice if 1.30 kg of the wood absorbs 6.75×104 joules of heat, and its temperatu
Gre4nikov [31]

Answer:

c=2.0769\  \frac{J}{g\ \textdegree C}

Explanation:

-Specific heat capacity is given by the formula:

q=mc\bigtriangleup T

Where:

q is the heat gained or loosed by the substance

m is the mass of the substance

c is the specific heat of the substance

\bigtriangleup T is the change in temperature

#We make c the subject of the formula and substitute to solve for it:

q=mc\bigtriangleup T\\\\c=\frac{q}{m\bigtriangleup T}\\\\\bigtriangleup T=(57-32)\textdegree C=25\textdegree C\\\\\therefore c=\frac{6.75\times 10^4J}{1.3\times 1000\ g\times 25\textdegree  C}\\\\=2.0769 \ \frac{J}{g\ \textdegree C}

Hence, the specific heat capacity of the ice is 2.0769 \frac{J}{g\ \textdegree C}

8 0
3 years ago
The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

3 0
3 years ago
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