All categories

3 years ago

How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC

Chemistry

1 answer:

Reptile [31]3 years ago

4 0

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

You might be interested in

11. If four gases in a cylinder each exert 6 atm, what is the total pressure exerted by the

Debora [2.8K]

Answer:

24 atm is the total pressure exerted by the gases

Explanation:

We propose this situation:

In a vessel, we have 4 gases (for example, hydrogen, Xe, methane and chlorine)

Each of the gases has the same pressure:

6 atm → hydrogen

6 atm → xenon

6 atm → methane

6 atm → chlorine

To determine the total pressure, we sum all of them:

Partial pressure H₂ + Partial pressure Xe + Partial pressure CH₄ + Partial pressure Cl₂ = Total P

6 atm + 6 atm + 6 atm + 6 atm = 24atm

8 0

4 years ago

Do you think the reflected waves would ever return to the airplane that transmitted them? explain

Ket [755]

No, it is very unlikely for that to happen.

8 0

3 years ago

Read 2 more answers

When NH3(g) reacts with N2O(g) to form N2(g) and H2O(g), 105 kcal of energy are evolved for each mole of NH3(g) that reacts. Wri

Ganezh [65]

Answer : The balanced chemical equation is,

$2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal$

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on product side.

The given unbalanced chemical reaction is,

$NH_3(g)+N_2O(g)\rightarrow N_2(g)+H_2O(g)+105kcal$

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of individual elements are not balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the $NH_3$ , the coefficient 3 is put before the $N_2O\text{ and }H_2O$ and the coefficient 4 is put before the $N_2$ .

The energy evolved in this reaction = $105Kcal\times 2=210Kcal$

Thus, the balanced chemical reaction will be,

$2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal$

7 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Two identical metal spheres at 8 °C are put into two beakers with equal amounts of water, as shown below.

leva [86]

Answer:

Correct answer is B.

Explanation:

Took the test and got this right. :)

7 0

3 years ago

Read 2 more answers