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3 years ago

Does friction always produce heat?

2 answers:

8 0

Friction produces heat whenever there’s surfaces going close to each other. It makes the molecules placed as quick and efficient energy towards it. I’m not really sure but perhaps?

AnnyKZ [126]3 years ago

5 0

Answer: Yes

Explanation: The objects rubbing together in opposite directions cause heat, which is friction.

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Show ur working:

AleksAgata [21]

Answer:

120 g SO3

Explanation:

128 g SO2 -> 160 g SO3

96 g SO2 -> x

x= (96 g SO2 * 160 g SO3)/128 g SO2

x= 120 g SO3

8 0

3 years ago

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr

jenyasd209 [6]

Answer : The final temperature of the water is, $22.5166^oC$

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

$\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole$

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives $\frac{6542 kJ}{2}\times 0.1103=360.7913kJ$ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : $q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat released = 360.7913 kJ = 36079.13 J

$m_w$ = mass of water = 5691 g

$c_w$ = specific heat of water= $4.18J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $21^oC$

Now put all the given values in the above formula, we get:

$36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)$

$T_{final}=22.5166^oC$

Therefore, the final temperature of the water is, $22.5166^oC$

8 0

4 years ago

What are bonds? <br> stocks <br> dividends <br> certificates of deposit <br> loans

tatuchka [14]

loans would be the answer

6 0

4 years ago

Read 2 more answers

Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butan

katrin2010 [14]

Answer:

a) pH will be 12.398

b) pH will be 4.82.

Explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

The total volume after addition of pure water = 0.780+0.01 = 0.79 L

The new concentration of /NaOH will be:

$molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M$

the [OH⁻] = 0.025

pOH = -log [OH⁻] = 1.602

pH = 14 -pOH = 12.398

b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

$pH=pKa+log\frac{[salt]}{[acid]}$

pKa= $-logKa=-log(1.5X10^{-5})=4.82$

ii) on addition of base the pH will increase.

8 0

4 years ago

The Questions are in the Pictures Attached

Whitepunk [10]

Answer:

1 is 5<3=me & you

Explanation:

7 0

3 years ago