Sedimentary rocks form ____

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

3 0

3 years ago

A microwave operates at 7.42 mm wavelength. How much energy is produced in Joules from 359 photons?

alisha [4.7K]

Answer:

$E=9.62\times 10^{-21}\ J$

Explanation:

Given that,

The wavelength of a microwave is 7.42 mm or 0.00742 m

No. of photons, n = 359

We need to find the energy produced by this no of photons. It can be given by the formula as follows :

$E=\dfrac{nhc}{\lambda}\\\\E=\dfrac{359\times 6.63\times 10^{-34}\times 3\times 10^8}{0.00742}\\\\=9.62\times 10^{-21}\ J$

or

$E=9.62e^{-21}\ J$

So, the required energy is $9.62\times 10^{-21}\ J$ .

3 0

3 years ago

Read 2 more answers

What did Mendel do that made his predictions so accurate

Hoochie [10]

He worked with large numbers of plants.

8 0

3 years ago

Read 2 more answers

How many moles of molecules are in 40g of H2O?

soldier1979 [14.2K]

Molecular mass of H2O is: 16+2*1=18
so moles of 40g of H2O is: 40/18=2.22

6 0

3 years ago

Read 2 more answers

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

3 years ago

Sedimentary rocks form _____.