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2 years ago

3

What is the amount of heat, in joules, required to increase the temperature of a 49.5-gram sample of water from 22°C to 66°C?

2 answers:

nalin [4]2 years ago

8 0

Answer:

(1) 9100 J.

Explanation:

This would be an increase of 44 degees C and the specific heat of water is 4.186 joules //g/degree C.

So it is 49.5 * 4.186 * 44

= 9100J.

pogonyaev2 years ago

6 0

Answer:

The answer to your question is: (1) 9100 J

Explanation:

Data

Q = ? J

mass = 49.5 g = 0.0495 kg

T1 = 22°C

T2 = 66°C

Cp = 4.18 J/Kg°C

Formula

Q = mCpΔT

Substitution

Q = (4.95)(4.18)(66 - 22)

Q = 9104 J

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Explanation:

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A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.)

Aliun [14]

The question is incomplete, here is the complete question:

A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.) Fill in the missing part of this equation.

$23.Pa.cm^3=?kPa.m^3$

<u>Answer:</u> The measurement after converting is $23\times 10^{-9}kPa.m^3$

<u>Explanation:</u>

We are given:

A quantity having value $23.Pa.cm^3$

To convert this into $kPa.m^3$ , we need to use the conversion factors:

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$1m^3=10^6cm^3$

Converting the quantity into $kPa.m^3$ , we get:

$\Rightarrow 23.Pa.cm^3\times (\frac{1kPa}{1000Pa})\times (\frac{1m^3}{10^6cm^3})\\\\\Rightarrow 23\times 10^{-9}kPa.m^3$

Hence, the measurement after converting is $23\times 10^{-9}kPa.m^3$

6 0

2 years ago

(a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compo

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Answer:

The molar mass of the compound is:- 168.82 g/mol

The molar mass of the gas is:- 16.38 g/mol

Explanation:

(a)

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRt$

Given that:-

Pressure = 20 kPa = 20000 Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

20000 Pa = $\frac {20000}{101325}$ atm

Pressure = 0.1974 atm

Temperature = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Molar mass = ?

Applying the equation as:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

<u>The molar mass of the compound is:- 168.82 g/mol</u>

(b)

Given that:

Pressure = 152 Torr

Temperature = 298 K

Volume = 250 cm³ = 0.25 L

Using ideal gas equation as:

$PV=nRT$

R = $62.3637\text{torr}mol^{-1}K^{-1}$

Applying the equation as:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Given that :

Mass of the gas = 33.5 mg = 0.0335 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.002045\ moles
= \frac{0.0335\ g}{Molar\ mass}$

<u>The molar mass of the gas is:- 16.38 g/mol</u>

5 0

2 years ago