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ollegr [7]
3 years ago
10

How will the rate at which a solid solute dissolves change if the solution is stirred?

Chemistry
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

The rate at which the solute dissolves will increase.

Explanation:

If a solution is stirred, the rate at which a solute dissolves would increase substantially provided the solution is not yet saturated.

Stiring would cause more of the solution to come in contact with every part of the solute. It will increase the surface area of contact for the solution to act which will shoot up the rate of reaction. Stiring helps to bring solutes in solutions into a more close contact with the molecules or compounds of the medium.

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A mixture of c2h2 and ch4 has a total mass of 230.9 g. when this mixture reacts completely with excess oxygen, the co2 and h2o p
Salsk061 [2.6K]
Interesting problem. Thanks for posting.

C2H2 + (3/2)02 ====>  H2O + 2CO2
CH4 +  2O2 =====> 2H2O + CO2

The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16

The number of moles of C2H2 = x
The number of moles of  CH4 = y
26x + 16y = 230.9 grams

For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18

x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water. 

Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2  we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2

Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y =  972.7

Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7

I'm not going to go through the math unless you request me to do so. 
x = 8.03 moles
y = 1.38 moles 

The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
6 0
2 years ago
Rank the following 0.100 M solutions in order of increasing H3O+ concentration:
Inessa05 [86]

Answer:

HCN < HOCl < HF  

Explanation:

The larger the Kₐ value, the stronger the acid.

6.2 × 10⁻¹⁰ < 4.0 × 10⁻⁸ < 6.3 × 10⁻⁴

   HCN      <     HOCl    <      HF

 weakest       stronger    strongest

5 0
3 years ago
How do I solve this?
Xelga [282]

Answer:

Explanation:

When you divide exponentials, you subtract the powers. For the numbers infront, just use a basic calculator for.

7.95/6.02 = 1.32

10^22/10^23 = 10^-1

1.32 x 10^-1 is your answer

7 0
3 years ago
Pls help me I don’t know what to dooooo
Ivan
Molarity = moles of solute/volume of solution in liters.

The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.

(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.

Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:

(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
5 0
3 years ago
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
2 years ago
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