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2 years ago

Ten pints of 15 % salt solution are mixed with 15 pints of 10 % salt solution. What is the

1 answer:

7 0

$Concentration\ rate=\frac{mass\ of\ product}{mass\ of\ substance}*100\%\\\\
15\%=\frac{m_1}{ms_1}*100\%\\ms_1=10pints\\
15\%=\frac{m_1}{10}*100\%\\
\frac{15}{100}=\frac{m_1}{10}\ \ |*10\\
m_1=\frac{150}{100}=1,5pints\\\\
10\%=\frac{m_2}{15}*100\%\\
0,1=\frac{m_2}{15}\ \ |*15\\
1,5=m_2\\
Resulting\ solution:\\
ms=ms_1+ms_2=15+10=25pints\\m=m_1+m_2=1,5+1,5=3pints
\\\frac{m}{ms}*100\%=\frac{3}{25}*100\%=12\%$

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1. Describe the difference between an electron and a proton in terms of their charge.

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Explanation:

Protons have a positive charge. Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.

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2 years ago

Which element will have the lower ionization energy? Be. Ba

sp2606 [1]

Boron would have the lower first ionization energy.
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3 years ago

Read 2 more answers

The average C - H bond energy in CH4 is 415 kJ / mol. Use the following data to calculate the average C - H bond energy in ethan

sertanlavr [38]

Answer and Explanation:

ΔH = (Bond energies of the products) - (Bond Energies of the reactants)

a) C2H6 + H2 ----> 2CH4 ΔH = -65.07 KJ/mol

Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol

Bond energy of reactants = Bond energy of C2H6 + Bond energy of H2

Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H)

Bond energy of H2 = (H-H) = 432 KJ/mol

-65.07 = 3320 - (Bond energy of C2H6 + 432)

Bond energy of C2H6 = 3320-432+65.07 = 2953.07 KJ/mol

Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H) = 2953.07

(C-H) = (2953.07 -347)/6 = 434.345 KJ/mol

b) C2H4 + 2H2 ------> CH4 ΔH = -202.21 KJ/mol

Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol

Bond energy of reactants = Bond energy of C2H4 + (2 × Bond energy of H2)

Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H)

(2 × Bond energy of H2) = 2 × 432 = 864 KJ/mol

-202.21 = 3320 - (Bond energy of C2H4 + 864)

Bond energy of C2H4 = 3320+202.21-864 = 2658.1 KJ/mol

Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H) = 2658.1

(C-H) = (2658.1 - 614)/4 = 511.05 KJ/mol

c) C2H2 + 3H2 ------> 2CH4 ΔH = -376.74 KJ/mol

Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol

Bond energy of reactants = Bond energy of C2H2 + (3 × Bond energy of H2)

Bond energy of C2H = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H)

(3 × Bond energy of H2) = 3 × 432 = 1296 KJ/mol

-376.74 = 3320 - (Bond energy of C2H2 + 1296)

Bond energy of C2H2 = 3320+376.74-1296 = 2400.74 KJ/mol

Bond energy of C2H2 = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H) = 2400.74

(C-H) = (2400.74 - 839)/2 = 780.87 KJ/mol

QED!!!

6 0

3 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago

Convert the following word equations into chemical equations using the correct

Alex17521 [72]

Answer:

Below.

Explanation:

1. Al2(SO4)^3 (aq) + 6LiOH aql) ----> 3Li2SO4 (aq) + 2Al(OH)^3 (s)

2. (NH4)2CO3 (aq) + MgCl2 (aq) ----> 2(NH4)Cl (aq) + MgCO3 (s).

7 0

3 years ago