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Y_Kistochka [10]
3 years ago
9

1. In a Stock system name such as iron(III) sulfate, the Roman numeral tells us (a) how many atoms of Fe are in one formula unit

. (b) how many sulfate ions can be attached to the iron atom. (c) the charge on each Fe ion. (d) the total positive charge of the formula unit. 2. Changing a subscript in a correctly written chemical formula (a) changes the number of moles represented by the formula. (b) changes the charges on the other ions in the compound. (c) changes the formula so that it no longer represents the compound it previously represented. (d) has no effect on the formula.
Chemistry
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

See explanation

Explanation:

1) The formula of the compound is Fe2(SO4)3. There are two ion atoms in any formula unit as we can see here. Three sulphate ions are attached to iron. Each iron ion carries a +3 charge as we saw in the formula. The total positive charge in each formula unit is +6.

Changing a subscript in a correctly written chemical formula changes the formula so that it no longer represents the compound it previously represented. Hence FeSO4 is a different compound from Fe2(SO4)3.

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74. Is the charge of a nucleus positive, negative, or zero?<br> The charge of an atom?<br> J
tia_tia [17]
The charge of a nucleus is neutral, meaning it’s zero

the positive charges of an atom equal the negative charges so the atom has a neutral charge
7 0
4 years ago
How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy
malfutka [58]

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01  moles of diethyl ether require heat = \frac{26.5}{1}\times 1.01=26.8kJ

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

5 0
4 years ago
What sort of nucleophile does NADH supply?
Elden [556K]

Answer:

Hydride ion

Explanation:

You often see the reduction with NADH written as

NADH ⟶ NAD⁺ + H⁺ + 2e⁻

If you think about it, H⁺ + 2e⁻ is equivalent to H:⁻, so we could write the reaction as

NADH ⟶ NAD⁺ + H:⁻

In terms of a mechanism, the dihydropyridine ring of NADH transfers a hydrogen atom with its pair of electrons (a hydride ion) to the substrate and becomes the more stable, aromatic pyridinium ion in NAD⁺.

7 0
4 years ago
Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16 0 mL of CO
aivan3 [116]

Answer:

V_{CO_2}=16.0mL

Explanation:

Hello,

In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:

n_{CO}V_{CO}=n_{CO_2}V_{CO_2}

Thus, solving for the yielded volume of carbon dioxide we obtain:

V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL

Best regards.

7 0
3 years ago
A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect
juin [17]

Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

b. The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

Kinetic energy = K = 4.25×10⁻⁵ J

a. What work was done by the electric force?

This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

V2 = -3.03 * 10³ V

The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

4 0
3 years ago
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