An atom has four electrons in its valence shell. What types of covalent bonds is it capable of forming?

The true statement is B. With identical masses for both metals, the final temperature of the two will be more aligned with 498 K rather than 298 K, as iron's specific heat capacity is significantly greater than that of gold's.

H2PO4- acts as a proton donor, whereas HPO42- serves as a proton acceptor. Step 1: Determining hydrogen ion donors and acceptors in the reaction displayed: H2PO4- is predisposed to release a H+ ion to yield HPO42-. On the other hand, HPO42- is inclined to accept a H+ ion, producing H2PO4-. The process of an acid in a water solvent is characterized as dissociation: HA ⇔ H+ + A- where HA denotes a proton acid. Therefore, H2PO4- = HA and HPO42- = A-. Acids are recognized as proton donors, which is why H2PO4- donates protons and HPO42- accepts them.

<span>Salts result from the reaction between bases and water. - FALSE
</span><span>Most salts are ionic and dissolve in water. - TRUE
</span><span>Most salts are not dissolved in water and do not have electrical charges. - FALSE
</span><span>Solutions containing salt and water are unable to conduct electricity. - FALSE

:)</span>

The visual representation is displayed in the following image.

For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; indicating the mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol / 2 mol CH3COOH)

= 0.005 mol

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm