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ra1l
1 month ago
13

What is the maximum volume of 0.25 M sodium hypochlorite solution (NaOCl, laundry bleach) that can be prepared by dilution of 1.

00 L of 0.80 M NaOCl?
Chemistry
2 answers:
KiRa [2.9K]1 month ago
8 0

Response:

0.313L

Clarification:

0.25Mx1.00L=0.80Mx V2

Thus, 0.25x1.00/0.80=0.313L

KiRa [2.9K]1 month ago
7 0

Answer: 3.2L

Explanation:

C1 = 0.80M

V1 = 1L

C2 = 0.25M

V2 =?

C1V1 = C2V2

0.8 x 1 = 0.25 x V2

V2 = (0.8 x 1) /0.25 = 3.2L

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Find the specific heat of sikver in J/g C if 38.5 cal is required to heat 25.0g of silver from 31.5 C to 58.7 C (1cal=4.184 J)
alisha [2963]

Answer:

The specific heat value for silver is 0.236 J/g-C.

Explanation:

Silver has a mass of 25 grams.

The temperature shifts from 31.5° C to 58.7° C.

The required heat is equivalent to 25 g.

To determine silver's specific heat, the following equation applies:

Q=mc\Delta T

Where c represents the specific heat of silver.

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{38.5 \times 4.184}{25(58.7-31.5)}\\\\c=0.236\ J/g-C

Thus, the specific heat of silver is 0.236 J/g-C.

7 0
1 month ago
26.2 g piece of copper metal is heated from 21.5°C to 201.6°C. Calculate the amount of heat absorbed by the metal. The specific
lorasvet [2795]

Response:1816.6 joules

Clarification:refer to the attached image

6 0
24 days ago
In basic solution, se2− and so32− ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this
lions [2927]

(a)   Write the balanced half-reactions for the overall process:

Oxidation: Se^2- (aq) → Se (s) + 2e-

Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- → S2O3^2- + 6OH- (aq)

(b)   Assuming E sulfite is 0.57 V, compute E selenium:

E anode = E cathode – E cell

= -0.57 – 0.35

= -.092

3 0
1 month ago
What is the percentage by mass of solution formed by dissolving 36.0 grams of HCl in 98.0 grams of water?
eduard [2782]
Convert HCl and H2O to moles.

36.0 g of HCl = 0.987 moles HCl

98.0 g of H2O = 5.44 moles H2O

Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.

For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.

Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O

Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.

11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O

11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.
4 0
1 month ago
Read 2 more answers
Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre
lions [2927]

Answer:

A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.

At this same depth, the density measures 2039 kg/m3.

Explanation:

P0 and ρ0 symbolize the pressure and density at sea level (indicative of atmospheric conditions). With an increase in ocean depth, both pressure and density likewise rise.

The relationship between pressure and density can be expressed as:

\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\

By rearranging

\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\

This equation allows for computation of P at 10,000 m beneath the ocean's surface:

P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa

The density found at a depth of 10,000 m in the ocean is

\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3

4 0
1 month ago
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