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3 years ago

Calculate the number of kilojoules to warm 125 grams of iron from 23.5 degrees Celcius to 78 degrees Celcius

1 answer:

3 0

Q=mcΔT

q=(125g)(0.446 J/(g x °C))(78°C - 23.5°C)

q=(125 g)(0.446 J/(g x °C))(54.5°C) (Be sure to cancel all similar units)

q= 3038.375 J

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Calculate [OH−] for each solution.

Triss [41]

Answer:

1. [OH⁻] = 0.30 M ; 2. [OH⁻] = 1.54x10⁻⁶M ; 3. [OH⁻] = 1.32x10⁻¹³M

Explanation:

Remember the rule:

pH + pOH = 14

pOH = 14 - pH

10*⁻pOH (you have to elevate 10, to -pOH)

10*⁻pOH = [OH⁻]

1. 14 - 13.48 = 0.52

10⁻⁰°⁵² = 0.30

2. 14 - 8.19 = 5.81

10⁻⁵°⁸¹ = 1.54x10⁻⁶

3. 14 - 2.12 = 12.88

10⁻¹²°⁸⁸ = 1.32x10⁻¹³

6 0

3 years ago

Which statements accurately describe elements? Check all that apply.

svet-max [94.6K]

Answer:

Elements are made up of only one type of atom.
Each element has a unique chemical symbol.
Elements can be identified by their atomic number.

Explanation:

7 0

2 years ago

Read 2 more answers

Assume that the maximum number of ATPs is produced (38). At pH 7, and in the presence of excess Mg2 , how much of the energy in

Lubov Fominskaja [6]

In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.

Explanation:

In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.

I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.

It can be said that only 38.3% of energy is put in ATP molecules.

8 0

3 years ago

Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of

Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

$\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles$

Now we have to calculate the moles of $PbO$

The balanced chemical reaction is,

$2Pb(s)+O_2(g)\rightarrow 2PbO(s)$

From the reaction, we conclude that

As, 2 mole of $Pb$ react to give 2 mole of $PbO$

So, 2.18 mole of $Pb$ react to give 2.18 mole of $PbO$

Now we have to calculate the mass of $PbO$

$\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO$

$\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g$

Theoretical yield of $PbO$ = 486.1 g

Experimental yield of $PbO$ = 367.5 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100$

$\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%$

Therefore, the percent yield of the reaction is, 75.6 %

3 0

3 years ago

A student performs three trials of a titration of an acid with an unknown concentration. She compares her measured concentration

Viktor [21]

Answer:

There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.

Main error that leads to the error in results is misreading of the end point volume .

End point is when the reaction between the analyte and solution of known concentration has stopped .

Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .

From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.

So, error may have occurred in wrongly judging of the end point by color change of the indicator .

4 0

3 years ago