Ask question

Ask question

All categories

3 years ago

10

Please help, need a correct answer asap!!

1 answer:

Tasya [4]3 years ago

4 0

CO2
hope this helps and brainly plez!

You might be interested in

What is the name of a solution whose concentration of solute is equal to the maximum concentration that

Travka [436]

Answer:

g

Explanation:

7 0

3 years ago

What is the answer to this?

Vesnalui [34]

Answer:

spring

Explanation:

cause i just took le test U-U

5 0

3 years ago

You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of

Karolina [17]

Answer:

37 mmol of acetate need to add to this solution.

Explanation:

Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-

$pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]$

Here pH is 5.31, $pK_{a}$ (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.

Plug in all the values in the above equation:

$5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]$

or, mmol of $CH_{3}COO^{-}$ = 37

So 37 mmol of acetate need to add to this solution.

3 0

2 years ago

What is the total number of moles represented by 20 grams of CACO3

Nutka1998 [239]

Answer:

B. 0.2.

Explanation:

We can use the relation:

<em>n = mass/molar mass</em>

mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.

<em>∴ n = mass/molar mass = </em>(20 g)/(100.0869 g/mol) <em>= 0.1998 ≅ 0.2 mol.</em>

<em></em>

<em>So, the right choice is: B. 0.2.</em>

8 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago