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irinina [24]
2 years ago
13

According to an informal 1992 survey, the drinking water in about one-third of the homes in Chicago had lead levels of about 10

ppb. Dr. Koether lived in Chicago from 1996 to 1998. Assuming she drank 1.4 L of water a day, calculate the total amount of lead in mg (using one decimal place) that she was exposed to over the two years if she lived in a home that had such high levels of lead.
Chemistry
1 answer:
wel2 years ago
8 0

Answer:

10.2 mg

Explanation:

Step 1: Calculate the total amount of water she drank

1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.

If she drank 1.4 L of water per day, the total amount of water she drank is:

730 day × 1.4 L/day = 1022 L

Step 2: Calculate the amount of Pb in 1022 L of water

The concentration of Pb is 10 ppb (10 μg/L).

1022 L × 10 μg/L = 10220 μg

Step 3: Convert 10220 μg to milligrams

We will use the conversion factor 1 mg = 1000 μg.

10220 μg × 1 mg/1000 μg = 10.2 mg

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Find the ratio v/cv/c for an electron in the first excited state (n = 2) of hydrogen.
dezoksy [38]

The answer is 0.365:100.

v/c ratio represents ratio of speed of an electron (v) to the speed of light (c).

How is the speed of an electron calculated?

  • The speed of an electron (v) is given by Bohr's model as-

v =\frac{1}{n}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h}

Now, for the first excited state, n = 2.

e - Charge of electron = 1.6×10^{-19} C

h - Plank's constant = 6.6×10^{-34} J.s

ε₀- permittivity = 8.85×10^{-12}N^{-1}.C^{2}.m^{-2}

  • Put the above data in the formula-

v =\frac{1}{2}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h}       =\frac{1}{4}\times \frac{(1.6\times 10^{-19})^{2}}{8.85\times 10^{-12}\times 6.6\times 10^{-34}}\\ \\     =0.01096\times 10^{8} \\      \\=1.096\times 10^{6}ms^{-1}

  • Now, the speed of light, c = 3.0×10^{8}\ m/s
  • Thus, the v/c ratio for an electron in the first excited state is calculated as-

\frac{v}{c} =\frac{1.096* 10^{6}\ m/s }{3.0 *10^{8}\ m/s }= \frac{0.365}{100}

  • Hence, the v/c ratio = 0.365:100.

To learn more about the speed of an electron (v), visit:

brainly.com/question/13198566

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6 0
1 year ago
For the reaction 4 FeCl2(aq) + 302(g) → 2Fe2O3(s) + 4Cl2(g), what volume of a
Darya [45]

Answer:

V=10.12mL

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible to evidence the 4:3 mole ratio between oxygen and iron (II) chloride; thus, we can compute the moles of the latter that are consumed by the given molecules of the former:

n_{FeCl_2}=4.32x10^{21}molec O_2*\frac{1molO_2}{6.022x10^{23}molec O_2} *\frac{4molFeCl_2}{3molO_2} \\\\n_{FeCl_2}=0.0095molFeCl_2

Now, since we have a 0.945-M solution of this iron (II) chloride, the corresponding volume turns out to be:

V=\frac{n_{FeCl_2}}{M}\\\\V=\frac{0.00956mol}{0.945mol/L}\\\\V=0.01012L\\\\V=10.12mL

Best regards!

5 0
3 years ago
Explain how water behaves in this reaction. Which definition of acids and bases would you apply?
mafiozo [28]

Water behaves as a base in this reaction.

The Bronsted-Lowry definition is applied, because the reaction involves the transfer of H+ from one reactant to the other.

A Bronsted-Lowry base is defined as a substance that accepts a proton.

Because water gains a proton to form H3O+ in this particular reaction, it acts as a base

5 0
2 years ago
Read 2 more answers
The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.
Nataliya [291]

a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = \frac{mass of element}{mass of compound}   (1)

The molecular formula of cisplatin is Pt(NH_3)_2Cl_2.

The atomic weight of the elements in cisplatin is:

Platinum, Pt = 195.084 u

Nitrogen, N = 14.0067 u

Hydrogen, H = 1 u

Chlorine, Cl = 35.453 u

The molar mass of Pt(NH_3)_2Cl_2 = 195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453) = 300.00 g/mol

The mass of each element calculated using formula (1):

- Platinum, Pt %

\frac{195.084}{300.00} \times 100 = 65.23%.

- Nitrogen, N%

\frac{2\times 14.0067}{300.00} \times 100 = 9.34%

- Hydrogen, H%

\frac{6\times 1}{300.00} \times 100 = 2.0%

- Chlorine, Cl%

\frac{2\times 35.453}{300.00} \times 100 = 23.63%

b. The given reaction of cisplatin is:

K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)

According to the balanced reaction, 1 mole of K_2PtCl_4 gives 1 mole of Pt(NH_3)_2Cl_2.

Now, calculating the number of moles of K_2PtCl_4 in 100.0 g.

Number of moles = \frac{given mass}{Molar mass}

Molar mass of K_2PtCl_4 = 2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol

Number of moles of K_2PtCl_4 = \frac{100 g}{415.093 g/mol} = 0.241 mole.

Since, 1 mole of K_2PtCl_4 gives 1 mole of Pt(NH_3)_2Cl_2. Therefore, mass of cisplatin is:

0.241 mole\times 300 g/mol = 72.3 g

For mass of KCl:

Molar mass of KCl = 39.0983 + 35.453 = 74.55 g/mol

Since, 1 mole of K_2PtCl_4 gives 2 mole of KCl. Therefore, mass of KCl is:

0.241 mole\times 74.55 g/mol\times 2 = 35.93 g


5 0
2 years ago
18 the first step in the mechanism is the acid catalyzed generation of an enol and then electrophilic addition of bromine. which
Aliun [14]

the first step in the mechanism is the acid-catalyzed generation of an enol and then electrophilic addition of bromine and cation is formed because of the destabilization effect of the electronegativity of oxygen

The ability of an atom or functional group to draw electrons to itself is known as an electronegativity in chemistry. An atom's electronegativity is influenced by both its atomic number and how far away from its charged nuclei its valence electrons are located.

The ability of an atom to draw shared electrons in a covalent connection is referred to as electronegativity. The stronger an element attracts the shared electrons, the higher its degree of electronegativity.

The propensity of an atom to attract other atoms when it is combined is known as an element's electronegativity. Additionally, a pair of bound electrons are shared. In contrast, an element's electropositivity refers to an atom's propensity to contribute electrons while also withdrawing from covalent connections.

To learn more about electronegativity please visit -
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5 0
2 years ago
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