What is the name given to the tiny clump of positive material at an atom’s center

Scilla [17]

Answer:

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$ .

Explanation:

Consider the oxidation state on each of the element:

Left-hand side:

O: -2 (as in most compounds);
Cr: $\displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6$ ;
Fe: +2 (from the charge of the ion);

Right-hand side:

Cr: +3;
Fe: +3.

Change in oxidation state:

Each Cr atom: decreases by 3 (reduction).
Each Fe atom: increases by 1 (oxidation).

Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.

Assume that the coefficient of the most complex species $\rm Cr_2O_7^{2-}$ is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.

Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain

two Cr atoms,
six Fe atoms, and
seven O atoms.

O atoms seldom appear among the products in acidic environments; they rapidly combine with $\rm H^{+}$ ions to produce water $\rm H_2O$ . Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen $\rm H^{+}$ ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$ .

7 0

3 years ago

Amantadine is effective in preventing infections caused by the influenza A virus and in treating established illnesses. It is th

Sliva [168]

Answer:

See image attached

Explanation:

a)

The full reaction mechanism of step 1 was obtained from Bartleby and attached to this answer. The steps involved in the reaction are:

1) Loss of Br- and formation of a carbocation

2) Attack of CH3CN on the carbocation

4) Formation of a quaternary nitrogen intermediate

5) Attack of water on the quaternary nitrogen intermediate

6) Loss of the water molecule

5) Formation of the amide product

b)

i) sodium hydroxide

ii) HCl

7 0

2 years ago

Which of the following could be one reason why the earth is getting warmer?

prisoha [69]

I think B. The Earth is moving closer to the sun
to me it makes more sense

8 0

2 years ago

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

lesya692 [45]

<u>Answer:</u> The nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

<u>Explanation:</u>

For the given element $_3^6\textrm{Li}$

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

$m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu$

M = mass of nucleus = $(n_p\times m_p)+(n_n\times m_n)$

$M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu$

Calculating mass defect of the nucleus:

$\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol$

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So, $0.032694g/mol=0.032694\times 10^{-3}kg/mol$

To calculate the nuclear binding energy, we use Einstein equation, which is:

$E=\Delta mc^2$

where,

E = Nuclear binding energy = ? J/mol

$\Delta m$ = Mass defect = $0.032694\times 10^{-3}kg/mol$

c = Speed of light = $2.9979\times 10^8m/s$

Putting values in above equation, we get:

$E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol$

Hence, the nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

3 0

3 years ago

Identify the molecule that is not a compound.

aev [14]

Compound is a molecule made of atoms from different elements. Allcompounds are molecules, but not allmolecules are compounds. Hydrogen gas (H2) is a molecule, but not a compound because it is made of only one element.

4 0

2 years ago