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likoan
4 months ago
8

Common additives to drinking water include elemental chlorine, chloride ions, and phosphate ions. Recently, reports of elevated

lead levels in drinking water have been reported in cities with pipes that contain lead, Pb(s). When Cl2(aq) flows through a metal pipe containing Pb(s), some of the lead atoms oxidize, losing two electrons each, and aqueous chloride ions form. (a) Write a balanced, net-ionic equation for the reaction between Pb(s) , and

Chemistry
1 answer:
alisha [2.9K]4 months ago
8 0

Answer:

Refer to the explanation

Explanation:

Please examine the image attached below for a detailed, step-by-step explanation regarding the above question.

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The density of an alcohol is 0.788 g/mL. What volume in microliters, μL, will correspond to a mass of 20.500 mg?
lions [2927]

Answer:

B.  26.0 μL.

Explanation:

Hello,

Considering the provided mass and density, the volume calculates to be:

V=\frac{m}{\rho} =\frac{25.000mg}{0.788g/mL}*\frac{1g}{1000mg}*\frac{1000\mu L}{1mL} \\ \\V=26.0\mu L

Thus, the solution is B.  26.0 μL.

Best regards.

6 0
3 months ago
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
3 months ago
The model below shows an atom of an element. 10 light gray and 8 dark gray balls sit at the center with 2 concentric black rings
Tems11 [2777]
The atomic number of the element indicated by the model is 8, corresponding to option 'C' in the question.
9 0
2 months ago
Read 2 more answers
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