All categories

3 years ago

A 40 g gold coin is cooled from 50°C to 10°C (CAu is 0.13 J/g-°

1 answer:

4 0

When the Heat gain or lose = the mass * specific heat * ΔT

and when we have the mass of gold coin= 40 g

and the specific Heat of gold= 0.13 J/g°

and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C

so by substitution:

∴Heat H = 40 g * 0.13 J/g° * -40
= - 208 J

You might be interested in

Explain the difference in the boiling point of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity an

Len [333]

The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.

As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.

The Interactions found in these compounds are London Dispersion Forces.

And among several factors at which London Dispersion Forces depends, one is the size of molecule.

Size of Molecule:
There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.

6 0

3 years ago

What element is least likely to be found in nature

Doss [256]

Answer:

uranium, caesium, potassium, beryllium,

Explanation:

5 0

3 years ago

In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones o

Advocard [28]

Answer:

52.54 %

Explanation:

Half life = 29 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{29.0}\ {years}^{-1}$

The rate constant, k = 0.023902 hour⁻¹

From 1964 to 1991:

Time = 27 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.023902\times 27}$

$\frac {[A_t]}{[A_0]}=0.5245$

<u>The strontium-90 remains in the bone = 52.54 %</u>

5 0

3 years ago

Can you guys help me please!!!!

Masteriza [31]

Answer:

i think it is true

Explanation:

if it correct plz plz mark as brainliest

thank you

19NBoli

4 0

3 years ago

Read 2 more answers

Which of the following statement is FALSE? a. Lowering the free energy of the transition state can increase a reaction rate. b.

ryzh [129]

Answer: Option (c) is the correct answer.

Explanation:

Activation energy or free energy of a transition state is defined as the minimum amount of energy required to by reactant molecules to undergo a chemical reaction.

So, when activation energy is decreased then molecules with lesser amount of energy can also participate in the reaction. This leads to an increase in rate of reaction.

Also, increase in temperature will help in increasing the rate of reaction.

Whereas at a given temperature, every molecule will have different energy because every molecule travels at different speed.

Hence, we can conclude that out of the given options false statement is that at a given temperature and time all molecules in a solution or a sample will have the same energy.

8 0

4 years ago