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2 years ago

Which of the following ionic compounds has a transition metal in it?

1 answer:

8 0

Considering that copper occupies Group 11, the ionic compound with a transition metal in it is c. CuSO₄.

<h3>What is a transition metal?</h3>

A transition metal is any of the set of metallic elements occupying a central block (Groups 3–12) in the periodic table. The differential electron occupies a d subshell.

Which of the following ionic compounds has a transition metal in it?

a. KOH. No, K occupies Group 1.
b. Cs₂S. No, Cs occupies Group 1.
c. CuSO₄. Yes, Cu occupies Group 11.
d. MgCl₂. No, Mg occupies Group 2.

The ionic compound with a transition metal in it is c. CuSO₄.

Learn more about transition metals here: brainly.com/question/2426896

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Please help me for question 1 and 2

likoan [24]

Answer:-

1) 6 mol

2) Mo

Explanation: -

Mass of Ozone = 48 g

Chemical formula of ozone = O3

Molar mass of Ozone O 3 = 16 x 3 = 48 g mol-1

Number of moles of ozone = Mass / molar mass

= 48 g / 48 g mol-1

= 1 mol

According to Avogadro’s law, 1 mole of a substance has 6.02 x 10^ 22 molecules.

So 1 mol of O3 has 6.02 x 10^ 22 molecules of ozone.

Now each Ozone molecule has 3 atoms of oxygen.

So, 1 mol of ozone has 3 x 6.02 x 10^22 atoms of oxygen.

Sodium must have 2 x 3 x 6.02 x 10^22 atoms as per the question.

According to Avogadro’s law, 6.02 x 10^ 22 atoms are in 1 mol of sodium

So, for 2 x 3 x 6.02 x 10^22 atoms, there should be (1/ 6.02 x 10^ 22) x 2 x 3 x 6.02 x 10^22

= 6 mol of sodium.

Let the mass of M be m g

Formula of hexafluoride = MF6.

Mass of the hexafluoride = g + 6 x 19

= m + 114

Mass of M=0.250g

Moles of M = 0.250/m

Mass of MF6= 0.547g

Moles of MF6 = 0.547/ (m + 114)

We know 1 mole of M gives 1 mole of MF6.

0.250/m moles of M gives 0.250/m moles of MF6.

But number of moles of MF6 = 0.547/ (m + 114)

Thus

0.250/m = (0.547)/ (m +114))

0.250m + 0.250 x 114 = 0.547m

m = 0.250 x 114 / (0.547 -0.250)

= 96

We see from the given data that Mo is 96.

So M is Mo.

4 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

3 years ago

What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO

docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

$\rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6$

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2$

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0

2 years ago

Which of the following solutions would have [Fe3+] = 0.020 M?

Shalnov [3]

Answer:

None of the given options

Explanation:

Let's go case by case:

A. No matter the volume, the concentration of Fe(NO₃)₃ (and thus of [Fe³⁺] as well) is 0.050 M.

B. We can calculate the moles of Fe₂(SO₄)₃:

0.020 M * 0.80 L = 0.016 mol Fe₂(SO₄)₃

Given that there are two Fe⁺³ moles per Fe₂(SO₄)₃ mol, in the solution we have 0.032 moles of Fe⁺³. With that information in mind we <u>can calculate [Fe⁺³]</u>:

0.032 mol Fe⁺³ / 0.80 L = 0.040 M

C. Analog to case A., the molar concentration of Fe⁺³ is 0.040 M.

D. Similar to cases A and C., [Fe⁺³] = 0.010 M.

Thus none of the given options would have [Fe⁺³] = 0.020 M.

7 0

2 years ago

Naturally occurring gallium is a mixture of isotopes

katrin2010 [14]

It is b have a great rest of your day

7 0

3 years ago