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kkurt [141]
3 years ago
10

149.3 g of H2O at 95 ◦C is poured over 412 g Fe at 5 ◦C in an insulated vessel. What is the final temperature? The specific heat

of H2O is 4.184 J /g· ◦C, the specific heat of Fe is 0.444 J/g · ◦C. Answer in units of ◦C.
I believe I am supposed to use the equation -Q=Q, where each Q=mc(change in T). But how do I know whether the water or the iron is the negative Q?
Chemistry
1 answer:
abruzzese [7]3 years ago
6 0
You are correct, but you needn't worry about the signs so much. Just remember that the negative sign is used to denote a loss of energy; since the water is hotter, it will be losing energy (-Q) and the iron will gain energy (Q). Now, we substitute the values:
-149.3 * 4.184 * (T - 95) = 412 * 0.44 * (T - 5)
Solving this equation for T,
T = 74.8 °C
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stellarik [79]

Answer:

Ka= [H+][But-]/H[Hbut]

Let k moles of H+ be formed

1.52*10^-5=k^2/0.004-k

K^2=0.004*1.52*10^-5-1.52*10^-5k

K^2- 1.52*10^-5k-6.08*10^-8=0

Solving this quadratic equation.

K=7.6*10^-6

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4 0
3 years ago
A 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.
dimaraw [331]

2.04 atm is the pressure of the nitrogen gas in the container when 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.

Explanation:

Data given:

moles of nitrogen gas, n = 0.316

volume of the nitrogen gas, V = 4 litres

temperature of the container of nitrogen gas  = 315 K

R (gas constant) = 0.08201 Latm/mole K

Pressure of the nitrogen gas on the container,  P = ?

from the data given, we will apply ideal gas law equation to calculate the pressure on the nitrogen gas:

PV = nRT

Rearranging the equation,

P = \frac{nRT}{V}

P = \frac{0.316 X 0.08201 X 315}{4}

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The pressure of the nitrogen gas in the container is 2.04 atm.

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The mineral manganosite, manganese(ll) oxide, crystallizes in the rock salt structure the face-centered structure adopted by NaC
balandron [24]

Answer:

A. 444.5 pm

Explanation:

We know that:

Density = \dfrac{mass \ of \ atoms \ in \ unit \ cell}{total \ volume \ of \ unit \ cell}

i.e.

\rho = \dfrac{n*M}{v_c * N_A}

\rho = \dfrac{n*M}{a^3 * N_A}

in a face-centered cubic crystal, the number of atoms per unit cell is (n) = 4

The molar mass of manganese (II) oxide [Mn(11)O] = 70.93 \ g/mol

Density \rho is given as 5.365 g/cm³

Avogadro constant N_A = 6.023 × 10²³ atoms/mol

∴

\rho = \dfrac{n*M}{a^3 * N_A}

Making th edge length "a" the subject, we get:

a^3 = \dfrac{n*M}{\rho* N_A}

a^3 = \dfrac{4*70.93 \ g/mol}{5.365 \ g/cm^3 *6.023 * 10^{23} \ atoms/mol }

a^3= 8.78 \times 10^{-23} \ cm^3

a= \sqrt[3]{8.78 \times 10^{-23} \ cm^3}

a = 4.445 × 10⁻⁸ cm

a = 444.5 pm

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Answer:

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