25.0g of iron is heated to 100.0 and then placed in 50.0 g of water in a insulated calorimeter. the initial temperature of the water is 38.00. the specific heat of water is 4.181j/g and the specific heat if iron is 0.45j/g. what is the final temp of the water and the iron?

Answers

Answer 1

Answer:

Approximately [tex]41.2\; {\rm ^{\circ} C}[/tex].

Explanation:

Let [tex]t\; {\rm ^{\circ} C}[/tex] be the final temperature of the water and the iron.

Temperature of the water would be increase by [tex](t - 38.00)\; {\rm ^{\circ} C}[/tex].

Temperature of the iron would be reduced by [tex](100.0 - t)\; {\rm ^{\circ} C}[/tex].

Let [tex]c[/tex] denote the specific heat of each material. Let [tex]m[/tex] denote the mass of the material. For a temperature change of [tex]\Delta t[/tex], the energy change involved would be:

[tex]Q = c\, m \, \Delta t[/tex].

The energy that the water need to absorb would be:

[tex]\begin{aligned}& Q(\text{water, absorbed}) \\ =\; & c(\text{water}) \, m(\text{water})\, \Delta t (\text{water}) \\ =\; & 4.181\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 50\; {\rm g} \times (t - 38.00)\; {\rm ^{\circ} C} \\ =\; & (209.05\, t - 7943.9)\; {\rm J} \end{aligned}[/tex].

The energy that the iron would need to release would be:

[tex]\begin{aligned}& Q(\text{iron, released}) \\ =\; & c(\text{iron}) \, m(\text{iron})\, \Delta t (\text{iron}) \\ =\; & 0.45\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 25.0\; {\rm g} \times (100.0 - t)\; {\rm ^{\circ} C} \\ =\; & (1125 - 11.25 \, t)\; {\rm J} \end{aligned}[/tex].

Since this calorimeter is insulated, the energy that the iron had released would be equal to the energy that the water had absorbed:

[tex]Q(\text{water, absorbed}) = Q(\text{iron, released})[/tex].

[tex]209.05\, t - 7943.9 = 1125 - 11.25\, t[/tex].

[tex]t \approx 41.2[/tex].

Thus, the final temperature of the water and the iron would be approximately [tex]41.2\; {\rm ^{\circ} C}[/tex].


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Iridium crystallizes in a face-centered cubic unit cell that has an edge length of 3.833 Å.
(a) Calculate the atomic radius of an iridium atom.
(b) Calculate the density of iridium metal

Answers

a.

The atomic radius of iridium is 1.355 Å

For a face-centered cubic unit cell, with edge length, a and atomic radius r,

we have a² + a² = (4r)²

2a² = 16r²

r² = a²/8

r = a/√8

Since a = 3.833 Å,

Substituting this into the equation, we have

r = a/√8

r = 3.833 Å/√8

r = 3.833 Å/2.828

r = 1.355 Å

So, the atomic radius of iridium is 1.355 Å

b.

The density of iridium metal is 22.67 g/cm³

To find the density of the iridium metal, we need to find the mass of atoms per unit cell.

So, for the face centered cubic iridium atom, there are 6 faces per unit cell × 1/2 atoms per face + 4 corners per unit cell × 1/4 atoms per corner = 3 atoms + 1 atom = 4 atoms per unit cell.

So, the mass of the atoms in the unit cell is m = number of atoms/cell × number of moles/atom × molar mass of Iridium

= 4 atoms/cell × 1 mol/6.022 × 10²³ atoms × 192.22 g/mol

= 768.88/6.022 × 10²³ g/cell

= 127.68 × 10⁻²³ g/cell

= 1.2768 × 10⁻²¹ g/cell

Next, we need to find the volume of the unit cell which is V = a³

= (3.833 Å)³

= (3.833 × 10⁻¹⁰ m)³

= 56.314 × 10⁻³⁰ m

= 5.6314 × 10⁻²⁹ m

So, the density of iridium = mass of unit cell/volume of unit cell

= m/V

= 1.2768 × 10⁻²¹ g/cell ÷ 5.6314 × 10⁻²⁹ m

= 0.2267 × 10⁸ g/m³

= 0.2267 × 10⁸ g/m³ × 10⁻⁶ m³/cm³

= 0.2267 × 10² g/cm³

= 22.67 g/cm³

So, the density of iridium metal is 22.67 g/cm³

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Answer:

[tex]FeCl_{2} +H_{2} S->FeS + 2HCl[/tex]

Explanation:

I am assuming that we have to balance this equation. On the left side, we have one Fe, 2 H, 2 Cl, and 1 S. On the right side, we have 1 Fe, 1 H, 1 Cl, and 1 S. Adding a 2 as a coefficient in front of the HCl on the right side will make 2 H and 2 Cl instead, balancing the overall equation.

FeCl_{2} +H_{2} S- > FeS + 2HClFeCl 2 +H 2 S−>FeS+2HCl

List THREE reasons why acidic soil can be a general problem to a natural ecosystem.

Answers

Answer:

Ok the poprtion would look like this :  x                   23  

                                                          -----------   =     ----------

                                                            230                 100

Explanation:

So for the last fill in it could be 52.9 or 60

i hope i got it correct lol pretty sure i did

Butane (C4 H10(g), Delta. Hf = â€"125. 6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Delta. Hf = â€"393. 5 kJ/mol ) and water (H2 O, Delta. Hf = â€"241. 82 kJ/mol) according to the equation below. 2 upper C subscript 4 upper H subscript 10 (g) plus 13 upper o subscript 2 (g) right arrow 8 upper C upper O subscript 2 (g 0 plus 10 upper H subscript 2 upper O (g). What is the enthalpy of combustion (per mole) of C4H10 (g)? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants. â€"2,657. 5 kJ/mol â€"5315. 0 kJ/mol â€"509. 7 kJ/mol â€"254. 8 kJ/mol.

Answers

The enthalpy of this reaction is -5315 KJ/mol.

The equation of the reaction is;

2C4H10(g) + 13O2(g) -----> 8CO2 (g) + 10H2O(g)

We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and products as follows;

ΔHrxn = ΔHf(products) - ΔHf(reactants)

We have the following information from the question;

ΔHf C4H10 = - 125. 6 kJ/mol

ΔHf CO2 = - 393. 5 kJ/mol

ΔHf H2O = - 241. 82 kJ/mol

ΔHf O2 = 0 KJ/mol

Hence;

[(8 × (- 393. 5 )) + (10 × (-  241. 82))] - [2( - 125. 6))]

= -5315 KJ/mol

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Is bleach liquid starch?
Yes or No

Answers

Answer:

O it's not

Explanation:

Have a great day!

It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reaction:


2 NaOH + H2CO3 → Na2CO3 + 2 H2O


The concentration of the carbonic acid is _____.

Answers

It requires 60 mL of 0.20 M sodium hydroxide [tex](NaOH)[/tex] to neutralize 25 mL of carbonic acid [tex](H_2CO_3)[/tex], hence the carbonate ions concentration is [tex]0.24M[/tex].

Given:

Reaction:

[tex]\to \bold{2NaOH + H_2CO_3 \to Na_2CO_3 + 2H_20 }[/tex]

[tex]NaOH[/tex] volume [tex](V_B) = 60 \ ml[/tex]  

[tex]H_2CO_3[/tex] Volume [tex](V_A) = 25\ ml[/tex]  

[tex]NaOH[/tex] Molarity [tex](C_B) = 0.20\ M[/tex]

[tex]H_2CO_3[/tex] moles [tex](n_A) = 1[/tex]

[tex]NaOH[/tex] moles [tex](n_B) = 2[/tex]

To find:

[tex]H_2CO_3[/tex] Molarity [tex](C_A) =?[/tex]

Solution:

Using the neutralization reaction:  

[tex]\to \frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B} \\\\[/tex]

[tex]\to C_B = 0.2\ M \\\\ \to n_A = 1 \\\\ \to n_B = 2 \\\\ \to V_B = 60\ ml \\\\ \to V_A = 25\ ml[/tex]

Calculating the [tex]C_A[/tex]:

 [tex]\to \frac{C_A \times 25}{0.2 \times 60} =\frac{1}{2} \\\\\to C_A =\frac{1 \times 0.2 \times 60}{2 \times 25} \\\\\to C_A =\frac{1 \times 2 \times 12}{2 \times 5 \times 10} \\\\ \to C_A =\frac{ 12}{ 5 \times 10} \\\\\to C_A = \frac{6}{5 \times 5} \\\\\to C_A = \frac{6}{25} \\\\\to C_A=0.24\ M[/tex]

Therefore, the concentration of carbonic acid is "0.24M".

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The concentration of carbonic acid( H₂CO₃) is 0.24 M.

The given reaction:

2 NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

Volume of NaOH  = 60mL

Volume of H₂CO₃ = 25 mL

Molarity of NaOH= 0.20M

To find:

Molarity of H₂CO₃=?

2 moles of NaOH and 1 mol of H₂CO₃ reacts to give 1 mol of Na₂CO₃.

Using the neutralization reaction:  

Consider A to be NaOH and B to be  H₂CO₃

[tex]\frac{\text{Number of moles of A}}{\text{Number of moles of B}} =\frac{\text{Molarity of A*Volume of A}}{\text{Molarity of B*Volume of B}}[/tex]

On substituting the values in order to calculate molarity of H₂CO₃:

[tex]\text{Molarity of B}=\frac{1*0.2*60}{2*25} \\\\\text{Molarity of B}=0.24M[/tex]

Therefore, the concentration of H₂CO₃ is 0.24 M.

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F
norm
app
F
grav
F
fric
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B. The size of Fnorm is the same as the size of Ffric
C. The size of Ffric is the same as the size of Fapp:
D. The size of F
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Fnorm = Fgrav

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Molecules are the smallest unit present in any compound and are formed by a group of atoms. A number denoted in front of the chemical species is the coefficient that represents the number of molecules.

The coefficient present in front of the reactant or the product in a chemical reaction gives the number of the molecules present in a compound. As three is denoted in front of the sulphuric acid, hence will be the number of the molecules.

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Which property determines whether a substance is a solid or liquid at room temperature?

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Answer:

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Answers

Answer:

Hey mate.....

Explanation:

This is ur answer.....

A triple bond is made of two pi bonds and one sigma bond. Examples of compounds with triple bonds include nitrogen gas, the cyanide ion, acetylene and carbon monoxide.

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Answer:

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Answer:

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