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Snowcat [4.5K]
2 years ago
15

I need anyone to answer me.. ​

Chemistry
1 answer:
artcher [175]2 years ago
4 0
A. 2,3
B. 19
C. 28
D. 33
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. Which laboratory equipment would most likely be used for testing a substance for the presence of monosaccharides? *
marishachu [46]

Answer:

Benedict's reagent is the indicator we use to detect monosaccharides. When monosaccharides are mixed with Benedict's and heated, a color ange occurs.

Hope this helps!! :)

3 0
3 years ago
Carbonic acid dissolves limestone and other rocks. This is an example of _____. chemical errosion
Jet001 [13]

Carbonic acid dissolves limestone and other rocks. This is an example of chemical erosion. An example is in the caves. Caves are formed where rainwater as it falls through the atmosphere absorbs carbon dioxide. The carbon dioxide makes the rain acidic to react it with the limestone bedrock. The rainwater is absorbed by the soil into the ground. Then as it enters through the soil, the rainwater will absorb more carbon dioxide that is produced by the decomposers. The carbon dioxide with water reacts to form carbonic acid. The carbonic acid will react to limestone and dissolves it slowly. As the space become larger, water can enter into it.

8 0
4 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
3 years ago
How can professionals increase their levels of stress as a result of personal work habits?
zmey [24]

Answer:

A

Explanation:

Developing work ethics won't lower your stress although it may make your conscience a little happier. Not B

Not C. Procrastinating just makes things harder. The mountain of things to do is not diminishing at all and your stress knows it.

Not D. Although D is the second best answer, setting goals is not really an action. It is a plan for how to work. But it is not work. Still having a goal is useful.

The answer is A. Just keep hitting what you have to do at a steady pace.

8 0
3 years ago
Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to
posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0
3 years ago
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