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Ymorist [56]
3 years ago
13

Sandra heated a 12.3-g piece of iron to 100oC. She then placed the iron in a calorimeter with a heat capacity of 10.4 J/oC. The

reading on the calorimeter temperature gauge rose from 25oC to 30oC. How much heat did the piece of iron lose?
A:     2.08 JB:     52.0 JC:     61.5 JD:     861 J
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0
We first have to find Q calorimeter

Q calorimeter = C (delta T) 

C= heat capacity in J/oC, not small c ( specific heat capacity) 
delta T= change in T (oC) = final T - initial T 

Q calorimeter = 10.4 J/oC (30 oC - 25 oC)
Q calorimeter = 52 J 

Remember, 

Q calorimeter = - Q surrounding
The (-) negative sign stands for heat being released. 

So , 

Q surrounding = - 52 J 

In other words, the iron releases 52 J .

The calorimeter absorbs the entire 52 J and heat rose from 25 - 30 oC. 

Answer: 52 J


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Bubba falls out of a plane flying over his peanut field. What will bubba speed be after falling for 8 seconds?
chubhunter [2.5K]

Variables we know:

t = 8 seconds

Vi = 0 m/s

g = -9.81

Δy = ?

Vf = ?

Equation we will be using to solve for Vf: Vf = Vi + gt

Steps to solve:

Vf = (0) + (-9.81)(8)

Vf = -78.48 m/s

Hope this helps!! :)

5 0
3 years ago
Read 2 more answers
Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo
stiv31 [10]
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: C_5H_{10}S_2
4 0
4 years ago
9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
3 years ago
Part 1: A cylinder containing 20.0 L of compressed nitrogen is connected to an empty (evacuated) vessel with an unknown volume.
baherus [9]

Answer:

The volume of the vessel is 250 L

Partial pressure of hydrogen = 189 torr

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 20.0 L

V₂ = ?

P₁ = 25 atm

P₂ = 2 atm

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{25}\times {20.0}={2}\times {V_2}

{V_2}=\frac {{25}\times {20.0}}{2}\ L

{V_2}=250\ L

<u>The volume of the vessel is 250 L.</u>

According to Dalton's law of partial pressure:-

P_{H_2}=Mole\ fraction\times Total\ Pressure

So, according to definition of mole fraction:

Mole\ fraction\ of\ H_2=\frac {n_{H_2}}{n_{H_2}+n_{He}}

Also,

Mole fraction of H₂ = 1 - Mole fraction of He = 1 - 0.75 = 0.25

So,

Total pressure = 756 torr

Thus,

P_{H_2}=0.25\times 756\ torr

<u>Partial pressure of hydrogen = 189 torr.</u>

7 0
3 years ago
Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and
andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454&#10;}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0
3 years ago
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