Ask question

Ask question

All categories

3 years ago

6

Identify the type of reaction: 2Ag + Cl2 --> 2AgCl

1 answer:

defon3 years ago

5 0

Answer:

The answer is E ( synthesis)

You might be interested in

A sealed piston contains 400 mL of air at standard ambient pressure. The piston is

nydimaria [60]

Answer:

324.24 kPa

Explanation:

Given that;

Initial pressure P1 = 101325 Pa

V1= 400 ml

P2 = ?

V2= 125mL

From Boyle's law;

P1V1 = P2V2

P2 = P1V1/V2

P2= 101325 × 400/125

P2= 324.24 kPa

5 0

3 years ago

Which of the following is a homogeneous mixture

anygoal [31]

Answer:

silicon or air or nickel or chocolate chip cookies i think

Explanation:

7 0

3 years ago

Can you give me a Biography about Mendeleev's Periodic Table?

Alecsey [184]

Answer: Dmitri Mendeleev was a Russian chemist who lived from 1834 to 1907. He is considered to be the most important contributor to the development of the periodic table. His version of the periodic table organized elements into rows according to their atomic mass and into columns based on chemical and physical properties

6 0

3 years ago

Rank these transition metal ions in order of decreasing number of unpaired electrons.

lesya692 [45]

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

Atomic number of Fe is 26.

$Fe^{3+} - [Ar] 3d^{5}$

So, there is only 1 unpaired electron present in $Fe^{3+}$ .

Atomic number of Mn is 25.

$Mn^{4+} - [Ar]3d^{3}$

So, there are only 3 unpaired electrons present in $Mn^{4+}$ .

Atomic number of V is 23.

$V^{3+} - [Ar] 3d^{2}$

So, there are only 2 unpaired electrons present in $V^{3+}$ .

Atomic number of Ni is 28.

$Ni^{2+} - [Ar] 3d^{8}$

So, there will be 2 unpaired electrons present in $Ni^{2+}$ .

Atomic number of Cu is 29.

$Cu^{+} - [Ar] 3d^{10}$

So, there is no unpaired electron present in $Cu^{+}$ .

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

7 0

3 years ago

What is the final temperature when 150.0 g of water at 90.0 °c is added to 100.0 g of water at 30.0 OC? Note that C, of water ca

levacccp [35]

Answer : The final temperature is, $337.8K$

Explanation :

$Q_{absorbed}=Q_{released}$

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 water at $90^oC$ = 150 g

$m_2$ = mass of water at $30^oC$ = 100 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of lead = $90^oC=273+90=363K$

$T_2$ = temperature of water = $30^oC=273+30=303K$

$c_1\text{ and }c_2$ = same (for water)

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

$150\times (T_{final}-363)=-[100\times (T_{final}-303)]$

$T_{final}=337.8K$

Therefore, the final temperature is, $337.8K$

8 0

4 years ago

Read 2 more answers