How many hydrogen atoms are in the following molecule of ammonium sulfide? (NH4)2S

Answer: The molecular formula will be

Explanation:

When percentages are provided, we assume the total mass to be 100 grams.

Thus, the mass of each element corresponds to the specified percentage.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1: convert given masses to moles.

Moles of C =

Moles of H =

Moles of O =

Step 2: For determining the mole ratio, divide each molar amount by the smallest number of moles calculated.

For C =

For H =

For O =

The resulting ratio of C: H: O= 4: 4: 1

Hence, the empirical formula obtained is

The empirical weight is calculated as = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now the molecular formula needs to be obtained.

The molecular formula can be derived as=

Answer:

CH4

Explanation:

The ideal behavior of gases generally depends on the strength of intermolecular forces between gas molecules and whether polar bonds are present.

In the case of CCl4, polar bonds exist along with the more electronegative chlorine atom, leading to stronger intermolecular forces at 400K, as opposed to CH4 which contains only non-polar bonds.

Thus, at 400K, CH4 behaves more like an ideal gas compared to CCl4.

Answer:

Calcium's atomic radius is roughly 175 pm.

Explanation:

We know that magnesium has an atomic radius of 150 pm.

The atomic radius of strontium measures 200 pm.

Since calcium's position is between magnesium and strontium in group 2 of the periodic table, its atomic radius should be roughly averaged between magnesium's and strontium's atomic radii because atomic radius is not constant.

Thus;

Calcium's atomic radius is approximately calculated as follows;

The average atomic radius is (200 + 150)/2 = 175 pm.

Answer:

C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.

Explanation:

Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.

This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.