Help me pls I will mark you as brain

A flask has a mass of 78.23g when empty and 593.63g when filled with water.When the same flask is filled with concentrateds ulfu

butalik [34]

Answer:

Density of concentrated H2SO4 = 1.99g/cm^3 = 1991.79Kg/m^3

Explanation:

mass of empty flask = 78.23g mass of flask filled when with water = 593.63g.

mass of flask filled when with concentrateds sulfuric acid, H2SO4 = 1026.57g

Mass of water = (mass of flask filled when with water) -

(mass of empty flask) = 593.63g - 78.23g = 515.4g

Volume of flask = volume of water = volume of concentrateds sulfuric acid, H2SO4 =

(mass of water)/ density of water) = 515.4g/1.00g/cm^3 = 515.4cm^3

The density of concentrated sulfuric acid is given by

Density of concentrated H2SO4 = (mass of H2SO4) ÷ (volume of H2SO4) = 1026.57g/515.4cm^3 = 1.99g/cm^3 = 1991.79Kg/m^3

7 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

4 years ago

Someone please help me please

igor_vitrenko [27]

Answer:

point 2 and 4

Explanation:

because at those point temperature is constant

3 0

3 years ago

Acids are and can destroy your skin and bones

sleet_krkn [62]

Answer:

This is true.

Explanation:

5 0

3 years ago

. In an acid-base titration, a standard solution of 1.0 MHCl is added to a solution of potassium hydroxide (KOH). As the standar

Hitman42 [59]

Answer:

The pH decreases very quickly, more quickly than earlier or later in the titration.

I think because SA-SB titration have the straight line where the equivalence point is where pH change is very steep and fast. Also pH is gonna decrease because after equivalence point you have more acid then base.

Explanation:

6 0

3 years ago