_____ are added to the soil to improve the fertility of the soil

If the specific heat of a solution is 4.18 J/goC, and you have 296 mL (1.03 g/mL) which increases in temperature by 6.9 degrees,

Sergeu [11.5K]

Answer:

Q = 8.8 kJ

Explanation:

Step 1: Data given

The specific heat of a solution = 4.18 J/g°C

Volume = 296 mL

Density = 1.03 g/mL

The temperature increases with 6.9 °C

Step 2: Calculate the mass of the solution

mass = density * volume

mass = 1.03 g/mL * 296 mL

mass = 304.88 grams

Step 3: Calculate the heat

Q = m*c*ΔT

⇒ with Q = the heat in Joules = TO BE DETERMINED

⇒ with m = the mass of the solution = 304.88 grams

⇒ with c = the specific heat of the solution = 4.18 J/g°C

⇒ with ΔT = the change in temperature = 6.9 °C

Q = 304.88 g * 4.18 J/g°c * 6.9 °C

Q = 8793.3 J = 8.8 kJ

Q = 8.8 kJ

8 0

3 years ago

Which is a difference between global and local winds?

valentina_108 [34]

Explanation:

local winds are considered breezes while global winds produce mostly storms

3 0

3 years ago

The elements of which group are nonreactive?

Margaret [11]

The elements of the "Noble" gases group is nonreactive. The reason for this is that noble gases are always or most of the time at room temperature.

3 0

3 years ago

Read 2 more answers

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate reacted = 0.2529 moles

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

Hence, the amount of oxygen gas collected will be 12.8675 g

5 0

3 years ago

What is a substance?

shusha [124]

I think the answer is C. An element or compound that cannot be physically separated. Sorry if im wrong.

7 0

4 years ago

_____ are added to the soil to improve the fertility of the soil​

_____ are added to the soil to improve the fertility of the soil