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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

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Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

3 years ago

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from

IgorLugansk [536]

The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g $S_8$ .

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of $S_8$ = 3.0 g

Molar mass of Ag = 107.8 g/mole

Molar mass of $S_8$ = 256 g/mole

Molar mass of $Ag_2S$ = 247.8 g/mole

First we have to calculate the moles of Ag and $S_8$ .

$\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles$

$\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)$

From the balanced reaction we conclude that

As, 16 mole of $Ag$ react with 1 mole of $S_8$

So, 0.0278 moles of $Ag$ react with $\frac{0.0278}{16}=0.00174$ moles of $S_8$

From this we conclude that, $S_8$ is an excess reagent because the given moles are greater than the required moles and $Ag$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag_2S$

From the reaction, we conclude that

As, 16 mole of $Ag$ react to give 8 mole of $Ag_2S$

So, 0.0278 moles of $Ag$ react to give $\frac{0.0278}{16}\times 8=0.0139$ moles of $Ag_2S$

Now we have to calculate the mass of $Ag_2S$

$\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S$

$\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g$

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

4 0

4 years ago

At standard pressure, the difference between the freezing point and the boiling point of water, in Kelvin degrees, is D. 373 A.

Sergio [31]

The answer to this question is A. 100

7 0

3 years ago

• Briefly discuss the cause of errors in the measurements

rewona [7]

(also called Observational Error) is the difference between a measured quantity and its true value. It includes random error

4 0

4 years ago

Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima

Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0

3 years ago