(34.969amu)(0.7577) = (36.966amu)(0.2423) = +

A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera

Grace [21]

Answer: The initial temperature of the iron was $515^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of iron = 360 g

$m_2$ = mass of water = 750 g

$T_{final}$ = final temperature = $46.7^0C$

$T_1$ = temperature of iron = ?

$T_2$ = temperature of water = $22.5^oC$

$c_1$ = specific heat of iron = $0.450J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]$

$T_i=515^0C$

Therefore, the initial temperature of the iron was $515^0C$

4 0

3 years ago

A sample of a gas in a rigid container has an initial pressure of 1.049 kPa and an initial temperature of 7.39 K. The temperatur

Doss [256]

Answer:- 4.36 kPa

Solution:- At constant volume, the pressure of the gas is directly proportional to the kelvin temperature.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Where the subscripts 1 and 2 are representing initial and final quantities.

From given data:

$P_1$ = 1.049 kPa

$P_2$ = ?

$T_1$ = 7.39 K

$T_2$ = 30.70 K

For final pressure, the equation could also be rearranged as:

$P_2=\frac{P_1T_2}{T_1}$

Let's plug in the values in it:

$P_2=\frac{1.049kPa(30.70K)}{7.39K}$

$P_2$ = 4.36 kPa

So, the new pressure of the gas is 4.36 kPa.

5 0

3 years ago

What is the empirical formula for a compound that contains 38.77% Cl and 61.23% O?

Leno4ka [110]

W(Cl)=0.3877
w(O)=0.6123
M(Cl)=35.5 g/mol
M(O)=16.0 g/mol

ClₐOₓ
M(ClₐOₓ)=35.5a+16.0x

w(Cl)=35.5a/(35.5a+16.0x)
w(O)=16.0x/(35.5a+16.0x)

solve a system of two equations with two unknowns
35.5a/(35.5a+16.0x)=0.3877
16.0x/(35.5a+16.0x)=0.6123

a=2
x=7

Cl₂O₇ is the empirical formula

3 0

3 years ago

Calculate the mass of iron(III) oxide that contains a million iron atoms. Be sure your answer has a unit symbol if necessary, an

KiRa [710]

Answer: The mass of iron (II) oxide that contains a million iron atoms is $2.6\times 10^{-17}g$

Explanation:

We are given:

Number of iron atoms = A million = $1.0\times 10^6$

The chemical formula of the given compound is $Fe_2O_3$

It is formed by the combination of 2 iron atoms and 3 oxygen atoms.

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

1 mole of iron (II) oxide will contain = $(2\times 6.022\times 10^{23})=1.2044\times 10^{24}$ number of iron atoms

We know that:

Molar mass of iron (II) oxide = 159.7 g/mol

Applying unitary method:

For $1.2044\times 10^{24}$ number of iron atoms, the mass of iron (II) oxide is 159.7 g

So, for $1.0\times 10^6$ number of iron atoms, the mass of iron (II) oxide will be $\frac{159.7}{1.2044\times 10^{24}}\times 1.0\times 10^6=2.6\times 10^{-17}g$

Hence, the mass of iron (II) oxide that contains a million iron atoms is $2.6\times 10^{-17}g$

8 0

3 years ago

PLEASE HELP fast (IMAGE)

n200080 [17]

Answer:

A. 3

Explanation:

There is half as many moles of CO₂ as of O₂. Half of 6 is 3. The equation is:

$6 molO2*\frac{1molCO2}{2molO2} = 3molCO2$

6 0

3 years ago