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2 years ago

Which of the following shows the correct rearrangement of the the heat equation q = mCpΔT to solve for specific heat?

2 answers:

6 0

When
q = m*Cp*ΔT

when q is Heat energy in Joules

and m is the mass of the substance in Kg

and Cp is the specific heat (J/Kg.K)

and Δ T is the change in temperature in Kelvin

so, by rearranging the formula we can get the specific heat Cp from:

∴Cp = q / m*ΔT

Kipish [7]2 years ago

5 0

Answer:

$C_{p} = \frac{q}{m(deltaT)}$

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Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected

Leviafan [203]

The question is incomplete; the complete question is;

Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set- up? Select one

a) There was an air bubble in the burette tip.

b) There is not enough indicator in the analyte.

c) The burette tip is leaking titrant into the analyte.

d) The analyte solution is being stirred too quickly

Answer:

a) There was an air bubble in the burette tip.

Explanation:

Titration involves the determination of the concentration of a solution by measuring the volumes of reactants used in the reaction. The concentration of one of the species must be known while the concentration of the other specie is to be determined by the volumetric analysis.

However, if there are air bubbles at the tip of the burette, this will cause less volume of titrant to be delivered from the burette than expected. Hence, the analyst may think that a certain volume of titrant has been delivered while in reality, a lesser volume was actually delivered due to the air bubbles present. Hence, the analyst may pass the expected endpoint without any colour change because of this problem.

8 0

1 year ago

Consider a buffer solution prepared from hocl and naocl. which is the net ionic equation for the reaction that occurs when naoh

marin [14]

OH⁻ from strong base (NaOH) react with weak acid from buffer (HOCl) according to the following equation:
OH⁻ + HOCl → H₂O + OCl⁻

6 0

2 years ago

Read 2 more answers

__ P4(s) + __ O2(g) → __ P4O10(s) Now we will balance O. How many O2 molecules are needed to form one P4O10 molecule?

Mnenie [13.5K]

Answer:

Five molecules of oxygen.

Explanation:

Hello,

In this case, considering the given chemical reaction, we must write a five before the oxygen in order to equal the number of oxygen atoms at both the right and left side (ten) so phosphorous remain the same (4):

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

It means that five molecules of oxygen (O₂) are needed to form one molecule of tetraphosphorous decaoxide (P₄O₁₀).

Best regards.

4 0

2 years ago

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p

xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

$Moles=1.20 \times {25.0\times 10^{-3}}\ moles$

Moles of potassium chloride = 0.03 moles

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=0.900 \times {15.0\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.0135 moles

According to the given reaction:

$2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

Theoretical yield = 3.75435 g

Given experimental yield = 2.45 g

% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %

6 0

2 years ago

Lithium ions in Lithium selenide (Li2Se) have an atomic radius of 73 pm whereas the selenium ion is 184 pm. This compound is mos

NARA [144]

Explanation:

Formula according to the radius ratio rule is as follows.

$\frac{r_{+}}{r_{-}} = \frac{73}{184}$

= 0.397

According to the radius ratio rule, as the calculated value is 0.397 and it lies in between 0.225 to 0.414. Therefore, it means that the type of void is tetrahedral.

Thus, we can conclude that the given compound is most likely to adopt closest-packed array with lithium ions occupying tetrahedral holes.

3 0

2 years ago