Identify the conjugate acid

Research the compositions of Pennies. What was the composition of each of your Pennies prior to treatment

LenKa [72]

Answer:

History of composition

Years Material Weight (grains)

1944–1946 gilding metal (95% copper, 5% zinc) 48 grains

1947–1962 bronze (95% copper, 5% tin and zinc) 48 grains

1962 – September 1982 gilding metal (95% copper, 5% zinc) 48 grains

October 1982 – present copper-plated zinc (97.5% zinc, 2.5% copper) 38.6 grains

7 0

2 years ago

I NEED AN ANSWER TO THESE QUESTIONS ASAP!! I GOT A TEST TOMORROW ON THIS!!!

Paha777 [63]

18.The octet rule tells us that in every chemical reactions, elements will either gain or lose electrons to attain the noble gas electron configuration. This stable electron configuration is known as the octet configuration since it is composed of 8 valence. Oxygen’s electron configuration is 1s2 2s2 2p4. So when oxygen reacts with other elements to form compounds, it completes the octet configuration by taking 2 electrons from the element it reacts with

19. Actually pure metals are made up not of metal atoms but rather of closely packed cations (positively charge particles). These cations are then surrounded by a pack of mobile valence electrons which drift from one part of the metal to another. This is called metallic bond.

20. This is the energy which is needed to break a single bond. When the dissociation energy is large, this means that the compound is more stable. Since carbon to carbon bonds have high dissociation energy, therefore they are not very reactive.

21. Network solids are type of solids in which the atoms are covalently bonded to one another, so they are very stable. It takes higher temperature to melt them because breaking these covalent bonds required greater energy. Some examples are:
- Diamond
-Silicon Carbide

5 0

3 years ago

Please help me with this!: Beverages such as pop and fizzy water are carbonated by dissolved CO2. Explain in detail why pop will

brilliants [131]

Answer:

If the cap is left off, some of the dissolved CO2 can escape as gas from the bottle, making the pop go flat faster (less dissolved CO2 in pop). If the cap is placed tightly, the gaseous CO2 cannot readily escape the bottle thus your pop won't go flat

Explanation:

If the cap is left off, some of the dissolved CO2 can escape as gas from the bottle, making the pop go flat faster. If the cap is placed tightly, the gaseous CO2 cannot readily escape the bottle thus your pop won't go flat.

Just some fun related concept:

A similar concept comes into play for the reason behind why pop tastes better in fridge then just keeping at normal temperature. This is because gases tend to have high solubility at cold temperatures thus CO2 is more readily dissolved in fridge than outside room temperature which is why it tastes great!

3 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

2 years ago

An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having

dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass 2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007 (1-x) = 1.008

2.014x + 1.007 - 1.007 x = 1.008

2.014x - 1.007x = 1.008 - 1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0

3 years ago