What are some of the reasons for arguing against adjusting what metals coins are made out of? (use textual evidence)

Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,

vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0

4 years ago

Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or nega

algol [13]

Answer: a) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$ : negative

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$ : positive

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$ : positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$ : negative

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$ : positive.

f) $I_2(s)\rightarrow I_2(g)$ : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$

As ions are moving to solid form , randomness decreases and thus sign of $\Delta S$ is negative.

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$

As gas is changing to solid, randomness decreases and thus sign of $\Delta S$ is negative.

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

f) $I_2(s)\rightarrow I_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

6 0

3 years ago

Consider the structures of vinegar and triglyceride and draw in any bond dipoles that exist in the molecules above using a dipol

bagirrra123 [75]

Answer:

Given molecules are vinegar and triglycerides.

Explanation:

The dipole is a vector quantity and it is heading from less electronegative atom to more electronegative atom in a polar covalent bond.

The structures and the bond dipoles in the given molecules are shown below:

5 0

3 years ago

Rank these real gases according to how closely they resemble an ideal gas. CO, N2, Ne, He, NH3

Gemiola [76]

He is the closest. Then:

Ne, N2, CO, NH3.

NH3 is the least closest to ideal.

8 0

3 years ago

Which type of flame(s) produce(s) incomplete combustion? Complete combustion?

Monica [59]

Answer:

See the answer and explanation below, please.

Explanation:

There are 2 types of combustion:

-Complete: C02 (carbon dioxide) is generated and the flame is blue, in the presence of a large amount of oxygen.

-Incomplete: it is also generated as a product C0 (carbon monoxide) harmful to health, and the color of the flame is yellow. It is produced in the presence of a small amount of oxygen.

8 0

3 years ago