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Maslowich
3 years ago
14

How much ice (in grams) would have to melt to lower the temperature of 354 ml of water from 26 ?c to 6 ?c? (assume the density o

f water is 1.0 g/ml.)?
Chemistry
1 answer:
Anit [1.1K]3 years ago
3 0
Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate things. We do as follows:

<span>Heat gained = Heat lost</span>
mC(T2-T1) = - mC(T2-T1) 

C(liquid water) = 4.18 J/gC
C(ice) = 2.11 J/gC

</span><span>(354 mL)(1.0 g/mL)(4.18 J/gC)(26 C - 6 C) = m(2.11 J/gC)(6 - 0C) </span><span>
m = 2337.63 g of ice

</span>
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A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4 and al
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Answer:

The value of K_p is 0.02495.

Explanation:

Initial concentration of SCL_2 gas = 0.675 M

Initial concentration of C_2H_4 gas = 0.973 M

Equilibrium concentration of mustard gas = 0.35 M

SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)

initially

0.675 M            0.973 M        0

At equilibrium ;

(0.675-0.35) M            (0.973-2 × 0.35) M        0.35 M

The equilibrium constant is given as :

K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}

=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}

K_c=14.45

The relation between K_p and K_c are :

K_p=K_c\times (RT)^{\Delta n}

where,

K_p = equilibrium constant at constant pressure = ?

K_c = equilibrium concentration constant =14.45

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

\Delta n = change in the number of moles of gas = [(1) - (1 + 2)]=-2

Now put all the given values in the above relation, we get:

K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}

K_p=6.2\times 10^{4}

K_p=0.02495

The value of K_p is 0.02495.

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2 years ago
A 25.0 mL sample of 0.100 M lactic acid (HC3H503, Ka = 1.4 E-4) is titrated with 0.100 M NaOH solution. Calculate the pH of the
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3 years ago
A sample of nitrogen gas is stored in a 489.6 mL flask at 108 kPa and 10.0°C. The gas is transferred to a 750.0 mL flask at 28.7
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Answer:

P2= 125.26 Kpa

Explanation:

V1= 489.6 ml=0.4896L

V2= 750 ml= 0.750L

V1= 180 KPa= 180000 Pa

P2= ?

T1= 10 = 10 + 273.15 = 283.15K

T2= 28.7+273.15= 301.85K

180000Pa* 0.4896L/ 283.15K * 301.85K/0.75L

P2= 12526.28553

P2= 125.26 KPa

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Answer:

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Answer:

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Explanation:

Ti is titanium and does not share any similar properties with the likes of Si, silicon and At, astatine.

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2 years ago
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