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frutty [35]
2 years ago
6

When the metal was placed in the calorimeter, its temperature dropped to the equilibrium temperature. Calculate the change in te

mperature of each metal sample. Record these values in Table B of your Student Guide. What is the change in temperature for the aluminum wire? C What is the change in temperature for the steel wire? C What is the change in temperature for the lead pellets? C
Chemistry
2 answers:
Murljashka [212]2 years ago
6 0

Answer:

1st one= -79.6

2nd one= -80

3rd one= -81.2

Explanation:

MAVERICK [17]2 years ago
5 0

Answer:

1st one= -79.6

2nd one= -80

3rd one= -81.2

Explanation:

Good luck!

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The number of sp2 hybrid orbitals on the carbon atom in CO32– is
Vladimir79 [104]
The number of sp2 hybrid orbitals on the carbon atom in CO32– is 3. Because hybrids = combination of 2 different types of orbitals
sp2 = 1/3 s character + 2/3 p character
7 0
2 years ago
Read 2 more answers
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K
Dmitry [639]

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

5 0
2 years ago
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
A 2135 cm3 sample of dry air has a pressure of 98.4 kpa at 127 degrees Celsius. What is the volume of the sample if the Temperat
kykrilka [37]

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2

2135 cm3 divided by 127 degrees celcius = x divided by 206

answer: 3460 cm3

7 0
2 years ago
Read 2 more answers
For a ternary solution at constant T and P, the composition dependence of molar property M is given by: M = x1M1 + x2M2 + x3M3 +
AveGali [126]

Answer:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i}) ...1

M^{\alpha } = M_{i} + CX_{xjxk}          ...2

Explanation:

The ternary constant is given by the following equation:

The symbol XiXi, where XX is an extensive property of a homogeneous mixture and the subscript ii identifies a constituent species of the mixture, denotes the partial molar quantity of species ii defined by

M_{i}  = [\frac{d(nM)}{dn_{i} }]_{P,t,n,j}

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In a multi phase system (in this case, a ternary system), the components resolved give:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i})

and M^{\alpha } = M_{i} + CX_{xjxk}

5 0
2 years ago
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