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2 years ago

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The lab procedure involves several factors, listed below. Some were variable and some were constant. Label each factor below V f

or variable or C for constant.
__ mass of the water in the calorimeter
__ mass of the metal
__change in temperature of the water
__change in temperature of the metal
__volume of water in calorimeter
__calorimeter pressure
__specific heat of water

2 answers:

Mariulka [41]2 years ago

8 0

Answer :

V - mass of the water in the calorimeter

V - mass of the metal

V - change in temperature of the water

V - change in temperature of the metal

C - volume of water in calorimeter

C - calorimeter pressure

C - specific heat of water

Explanation :

Variables : It is a factor that changes during the experiment or calculation.

Constant : It is a factor that does not change during the experiment or calculation.

In a calorimeter, the heat absorbed is equal to the heat released.

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T$

$m_1\times c_1\times \Delta T_1=-[m_2\times c_2\times \Delta T_2]$

where,

$m_1$ = mass of water in calorimeter

$m_2$ = mass of metal

$\Delta T_1$ = change in temperature of the water

$\Delta T_2$ = change in temperature of the metal

$c_1$ = specific heat of water

$c_2$ = specific heat of metal

From this, we conclude that the value of specific heat of water is constant while the other are variables.

The volume of water in calorimeter, calorimeter pressure is also constant.

Lesechka [4]2 years ago

7 0

Answer:

v - mass of the water in the calorimeter

v - mass of the metal

v - change in temperature of the water

v - change in temperature of the metal

c - volume of water in calorimeter

c - calorimeter pressure

c - specific heat of water

Explanation:

just did it on edge

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Answer:

The correct answer is "1.0100".

Explanation:

Let the volume of mixture be 100 ml.

then,

The volume of DMSO will be 10 mL as well as that of water will be 90 mL.

DMSO will be:

= $10\times 1.1004$

= $11.004 \ g$

The total mass of mixture will be:

= $90+11.004$

= $101.004 \ g$

Density of mixture will be:

= $\frac{Mass}{Volume}$

= $\frac{101.004}{100}$

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hence,

Specific gravity of mixture will be:

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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 =

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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:

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Mass of disodium succinate:

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

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A scientist compares two samples of white powder. one powder was present at the beginning of an experiment. the other powder was

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As density is an intensive property of the substance.Which means that different substance have different densities.

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On comparing both the densities of the powders we can say that both the substances are different. So we can conclude that the chemical reaction had occurred.

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Answer:

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Explanation:

The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.

The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.

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Explanation:

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