In the Minnesota Department of Health set a health risk limit for benzene in groundwater of . Suppose an analytical chemist receives a sample of groundwater with a measured volume of . Calculate the maximum mass in milligrams of benzene which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits.

Answers

Answer 1

Answer:

3.10 × 10⁻⁴ mg

Explanation:

In the Minnesota Department of Health set a health risk limit for benzene in groundwater of 10.0 μg/L. Suppose an analytical chemist receives a sample of groundwater with a measured volume of 31.0 mL. Calculate the maximum mass in milligrams of benzene which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Concentration limit (C): 10.0 μg/LVolume of the sample (V): 31.0 mL

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

31.0 mL × 1 L/1000 mL = 0.0310 L

Step 3: Calculate the maximum mass of benzene (m)

We will use the following expression.

m = C × V = 10.0 μg/L × 0.0310 L = 0.310 μg.

Step 4: Convert "m" to mg

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

0.310 μg × 1 mg/1000 μg = 3.10 × 10⁻⁴ mg


Related Questions

A chemist dissolves of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The temperature of the solution is .) Be sure your answer has the correct number of significant digits.

Answers

Answer:

12.99

Explanation:

A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.

Step 1: Given data

Mass of KOH: 716. mg (0.716 g)Volume of the solution: 130. mL (0.130 L)

Step 2: Calculate the moles corresponding to 0.716 g of KOH

The molar mass of KOH is 56.11 g/mol.

0.716 g × 1 mol/56.11 g = 0.0128 mol

Step 3: Calculate the molar concentration of KOH

[KOH] = 0.0128 mol/0.130 L = 0.0985 M

Step 4: Write the ionization reaction of KOH

KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)

The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

Step 5: Calculate the pOH

We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

Step 6: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -1.01 = 12.99

how to remove acetone from a school chromebook

Answers

Answer:

Apply some dishwashing solution with a white washcloth to the acetone residue and for a few minutes, work in gently from the edges of the spill to the center. Dont rub. Continue until the spill is completely removed.

Answer:

Blot the liquid with a dry, white absorbent cloth. Gently scrape up semi-solids with a rounded spoon. Break up solids and use something to remove them

What do we call the study of weather? A.forcasting B.meteorology​

Answers

The correct answer is B. meteorology. The study of weather is called meteorology, derived from the Greek word for "of the atmosphere".

PLEASE HELP it would be really nice of you to answer this question i would be very grateful <3

If an element has three isotopes with known masses, what other information
is needed to find the average atomic mass of the element?
A. The number of neutrons in each isotope
B. The average mass of the isotopes
C. The atomic number of each isotope
D. The abundance percentage of each isotope

Answers

Answer: D (The abundance percentage of each isotope)

Explanation: hope this helps!

The answer is pretty much d

I need help on this question

Answers

Answer:

warm blooded

Explanation:

humans are warm blooded. we cannot change our body temperature as we want to

do you still need help ?

Gallium (Ga) consists of two naturally occurring isotopes with masses of 68.926 and 70.925 amu. (a) How many protons and neutrons are in the nucleus of each isotope

Answers

Answer:

[tex]neutrons_{Ga-69}=38\\\\neutrons_{Ga-71}=40[/tex]

Explanation:

Hello!

In this case, since isotopes of the same element have the same number of protons and electrons but different atomic mass, we can compute the number of neutrons by subtracting the number of protons to the atomic mass of the isotope; thus, for Ga-69 and Ga-71 (rounded up to whole numbers), we obtain:

[tex]neutrons_{Ga-69}=69-31=38\\\\neutrons_{Ga-71}=71-31=40[/tex]

Best regards!

Select the atoms that belong to the same element.

Answers

Answer:

the middle and bottom left

Explanation:

one is a positively charged atom, and the other is normal. they are both the same

Answer:

the middle and bottom left .

Severe or sudden changes in Earth's atmosphere can cause entire species of animals to die. When all of the members of species die out, it is called

Answers

Answer: Extinction

Animals on the verge of extinction are considered endangered species. Some could argue that the phrasing "threatened" is synonymous to "endangered". Some classify extinction based on if the species is completely wiped out in the wild, or if the species is completely gone (both in the wild and in zoos for instance). Extinction events are not solely due to the changes in atmosphere, but also due to lack of food or habitat, among other possible factors.



HELP
How many grams are in 2.500 mol of magnesium hydroxide?

58.33 g
0.04286 g
145.8 g
100.8 g
23.33 g

Answers

Answer:

145.8g

Explanation:

Given parameters:

Number of moles of magnesium hydroxide  = 2.5mol

Unknown:

Mass of Mg(OH)₂  = ?

Solution:

To solve this problem we use the expression below;

  Mass of Mg(OH)₂ = number of moles x molar mass

 Molar mass of Mg(OH)₂ = 24.3 + 2(16 + 1)  = 58.3g/mol

 Mass of Mg(OH)₂ = 2.5 x 58.3  = 145.8g

The vapor pressure of ethanol, CH3 CH2 OH, at 40.0 °C is 17.88 kPa. If 2.24 g of ethanol is enclosed in a 3.00 L container, how much liquid will be present?​

Answers

Answer:

7.13 L of ethanol, CH₃CH₂OH

Explanation:

We'll begin by calculating the number of mole in 2.24 g of ethanol. This can be obtained as follow:

Mass of CH₃CH₂OH = 2.24 g

Molar mass of CH₃CH₂OH = 12 + (3×1) + 12 + (2×1) + 16 + 1

= 12 + 3 + 12 + 2 + 16 + 1

= 46 g/mol

Mole of CH₃CH₂OH =?

Mole = mass / molar mass

Mole of CH₃CH₂OH = 2.24 / 46

Mole of CH₃CH₂OH = 0.049 mole

Next, we shall convert 40 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 40 °C

T(K) = 40 °C + 273

T(K) = 313 K

Finally, we shall determine the volume of ethanol, CH₃CH₂OH in present in the container. This can be obtained as follow:

Pressure (P) = 17.88 KPa

Temperature (T) = 313 K

Number of mole (n) = 0.049 mole

Gas constant (R) = 8.314 L.KPa/Kmol

Volume (V) =?

PV = nRT

17.88 × V = 0.049 × 8.314 × 313

Divide both side by 17.88

V = (0.049 × 8.314 × 313) / 17.88

V = 7.13 L

Thus, 7.13 L of ethanol, CH₃CH₂OH will be present in the container.

Based on the mass of ethanol provided, the mass of liquid ethanol is 1.61 g

What is the relationship between gas volume, pressure, temperature and moles?

The relationship between the gas volume, pressure, temperature and moles is given by the ideal gas equation:

PV = nRT

where:

P is pressureV is volumen is number of molesR is molar gas constant = 8.314 L.KPa/KmolT is temperature

From data provided:

P =  17.88 kPa

V = 3.00 L

T = 40.0 °C = 313 K

R = 0.049 mole

n = ?

n = PV/RT

n =  17.88 * 2/8.314  * 313

n = 0.0137 moles

Mass of ethanol gas = number of moles * molar mass

Mass of gaseous ethanol = 0.0137 mole * 46 g/mol

Mass of gaseous ethanol = 0.63 g

Then:

Mass of liquid ethanol, CH₃CH₂OH = 2.24 - 0.63

Mass of liquid ethanol, CH₃CH₂OH = 1.61 g

Therefore, the mass of liquid ethanol is 1.61 g

Learn more about vapor pressure and mass at:

if i have 4 moles of a gas at a pressure of 5 atm and a volume of 12 liters what is the temperature?

Answers

Answer:

182.70K

Explanation:

Using the general gas equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = 0.0821 Latm/Kmol

T = temperature (K)

Based on the provided information, P = 5 atm, V = 12L, n = 4 moles, T =?

PV = nRT

5 × 12 = 4 × 0.0821 × T

60 = 0.3284T

T = 60/0.3284

T = 182.70K

if you are provided with 34.5 g iron (iii) chloride and 25.3 g sodium carbonate what is the limiting ractant

Answers

Answer:

Sodium carbonate, Na₂CO₃ is the limiting

Explanation:

We'll begin by writing the balanced equation for the reaction between iron (iii) chloride, FeCl₃ and sodium carbonate, Na₂CO₃.

This is illustrated below:

2FeCl₃ + 3Na₂CO₃ → Fe₂(CO₃)₃ + 6NaCl

Next, we shall determine the masses of FeCl₃ and Na₂CO₃ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of FeCl₃ = 56 + (3×35.5)

= 56 + 106.5

= 162.5 g/mol

Mass of FeCl₃ from the balanced equation = 2 × 162.5 = 325 g

Molar mass of Na₂CO₃ = (2×23) + 12 + (3×16)

= 46 + 12 + 48

= 106 g/mol

Mass of Na₂CO₃ from the balanced equation = 3 × 106 = 318 g

SUMMARY:

From the balanced equation above,

325 g of FeCl₃ reacted with 318 g of Na₂CO₃.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

325 g of FeCl₃ reacted with 318 g of Na₂CO₃.

Therefore, 34.5 g of FeCl₃ will react with = (34.5 × 318)/325 = 33.76 g of Na₂CO₃.

From the calculations made above, we can see clearly that it will take a higher amount (i.e 33.76 g) of Na₂CO₃ than what was given (i.e 25.3 g) to react completely with 34.5 g of FeCl₃. Therefore, Na₂CO₃ is the limiting reactant and FeCl₃ is the excess reactant.

what are three components that make a complete description of a location?

Answers

Answer:

Distance, direction and symbol.

Explanation:

Distance, direction and symbol are the three components of the map which make a complete description of a location. On the map, these three components are present which provides information about a specific location on the map. Map is a drawing on the paper that shows the geography of the whole world and provides information in detail.

How does ocean temperatures very around the globe?


PLZ HELP ME

Answers

Answer:

As the oceans absorb more heat, sea surface temperature increases and the ocean circulation patterns that transport warm and cold water around the globe change

Explanation:

3. Which type of pitch do you think was
thrown in the following situation?
A pitcher throws a pitch that looks like
a fastball but travels a little slower. The
batter swings too early, expecting a
fastball.
A. Change-up
B. Slider
C. Fastball
D. Curveball

Answers

the answer to your question is ‘B

Write down the formulas of the following compounds: magnesium chloride, alumina, chlorine (VII) oxide, sodium sulfide, calcium oxide.

Answers

Answer:

- Magnesium chloride: MgCl₂.

- Alumina: Al₂O₃.

- Chlorine (VII) oxide: Cl₂O₇.

- Sodium sulfide: Na₂S.

- Calcium oxide: CaO.

Explanation:

Hello!

In this case, according to the naming rules for the given compounds, which state that binary salts have a metal and a nonmetal and oxides contain oxygen, we have:

- Magnesium chloride: MgCl₂.

- Alumina: Al₂O₃.

- Chlorine (VII) oxide: Cl₂O₇.

- Sodium sulfide: Na₂S.

- Calcium oxide: CaO.

Best regards!

If data comes in that disproves a theory, what will happen to the theory?​

Answers

Answer: More conservative scientists will try to modify the existing theory to fit the new data. The last answer, the fourth one.

Explanation: I think I had this question before, hope it helps!

ANSWER FAST PLS>> WILL GIVE BRAINLIEST

A 52.6 kg car is driving at a constant velocity of 20 m/s over 2 seconds. What is its acceleration?

Answers

Answer:

Centripetal acceleration is related to tangential velocity through ac = v ^ 2 / R

The angular velocity is related to the tangential velocity through ω = v / R

a) ac = (20 m / s) ^ 2/20 m = 20 m / s ^ 2

b) ω = 20 m / s / 20m = 1 rad / s

c) The swept angle is Ф = ω.t = 1 rad / s. 3600 s = 3600 rad

Expressed in turns it is Ф = 3600 rad. 1 turn / (2.π rad) = 573 turn

Explanation:

In what element will metallic bonding occur?
A.
Carbon
B.
Oxygen
C.
Argon
D.
Cobalt

Answers

D




A- Harvard university professor

What does the interquartile range represent?
A
the lower 50 percent of the data

B
the higher 50 percent of the data

C
the middle 50 percent of the data

D
the middle data point

Answers

Answer: C) middle 50 percent of the data

The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.

25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.

The middle 50 percent of the data. Therefore, option (C) is correct.

The interquartile range (IQR) shows the spread or dispersion of a dataset. It is calculated as the difference between the third quartile (Q3) and the first quartile (Q1) in a sorted dataset. The first quartile (Q1) is the value below which 25 percent of the data falls, and the third quartile (Q3) is the value below which 75 percent of the data falls.

The interquartile range captures the range of the central 50 percent of the data. It provides a measure of the variability or spread of the middle portion of the dataset while excluding the influence of extreme values or outliers.

Learn more about interquartile range, here:

https://brainly.com/question/29173399

#SPJ6

In 42 hrs, 4.32 g of argon-35 decay to leave only 0.54 g. What is the half-life of argon-

35?

o 21 hrs

O 7 hrs

O 84 hrs

14 hrs

Answers

Answer:

14 hrs.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 42 hrs

Original amount (N₀) = 4.32 g

Amount remaining (N) = 0.54 g

Half-life (t½) =…?

Next, we shall determine number of half-lives required for 4.32 g of the isotope to decay to 0.54 g. This can be obtained as follow:

Original amount (N₀) = 4.32 g

Amount remaining (N) = 0.54 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

0.54 = 1/2ⁿ × 4.32

Cross multiply

0.54 × 2ⁿ = 4.32

Divide both side by 0.54

2ⁿ = 4.32 / 0.54

2ⁿ = 8

Express 8 in index form with 2 as the base

2ⁿ = 2³

n = 3

Thus, it took 3 half lives for 4.32 g of the isotope to decay to 0.54 g.

Finally, we shall determine the half-life of the isotope. This can be obtained as follow:

Time (t) = 42 h

Number of half-lives (n) = 3

Half-life (t½) =

n × t½ = t

3 × t½ = 42

Divide both side by 3

t½ = 42 / 3

t½ = 14 hrs

Thus, the half-life of the isotope is 14 hrs

how many electrons does cesium gain or lose

Answers

cesium : [Xe]6s1 [ X e ] 6 s 1 . It loses 1 electron from 6s orbital and attains the noble gas configuration of Xenon.

Because of the magnetic field, magnets do have to touch to affect each other. Question 6 options: True False

Answers

Answer:

False, they can interact without touching

1
What is an example of a phenotype?
A having dark hair
B. having dominant genes
C. having recessive genes
D. SS

Answers

I believe that the answer is B
The correct answer is A

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a con-
stant extemal pressure of 15 atm.

Answers

Answer:

QhhjjgDjKfxockzgnx

Explanation:

DjiDfj,jzo

The radius of a gold atom is 144 pm. How many gold atoms would have to be laid side by side to span a distance of 4.97 mm

Answers

Answer:

2.18 × 10⁷ Au atoms

Explanation:

Step 1: Calculate the diameter of a gold atom

The diameter is twice the radius.

d = 2 × r = 2 × 144 pm = 228 pm

Step 2: Convert "d" to meters

We will use the conversion factor 1 m = 10¹² pm.

228 pm × 1 m/10¹² pm = 2.28 × 10⁻¹⁰ m

Step 3: Convert the total distance to be covered (D) to meters

We will use the conversion factor 1 m = 10³ mm.

4.97 mm × 1 m/10³ mm = 4.97 mm × 10⁻³ m

Step 4: Calculate the number of atoms of gold required

We will divide the total distance by the diameter of 1 atom of gold.

D/d = 4.97 mm × 10⁻³ m/2.28 × 10⁻¹⁰ m = 2.18 × 10⁷

You dissolve 15.71 g of NH4NO3 in 150.0 mL of solution (call this solution A). You take 20.0 mL of solution A and add water until the total volume is 75.0 mL (call this solution B). You take 15.0 mL of solution B and add 25.0 mL of water to it (call this solution C). You mix 10.0 mL of solution B and 10.0 mL of solution C (call this solution D). What is the concentration of ammonium nitrate in solution D

Answers

Answer:

[NH₄NO₃] at D → 0.279 M

Explanation:

This exercise involves a series of dilutions one after the other.

First of all, we calcualte ammonium nitrate's concentration at A.

15.71 g . 1 mol/ 80 g = 0.196 mol / 0.150 mL = 1.31 M

At B → 1.31 M . 20 mL/ 75 mL = 0.349  M

At C →  0.349 M . 15 mL / 25 mL = 0.209 M

[NH₄NO₃] at B = 0.349 M

[NH₄NO₃] at C = 0.209 M

So let's calculate the new moles

In 1 mL of B we have 0.349 mmoles

In 10 mL of B we have 3.49 mmoles

In 1 mL of C we have 0.209 mmoles

In 10 mL of C we have 2.09 mmoles

Volume of D = 10 ml + 10ml = 20 mL

Total mmmoles = 3.49 mmoles + 2.09 mmoles = 5.58 mmoles

[NH₄NO₃] at D = 5.58 mmoles / 20mL → 0.279 M

At rest, a neuron has a lower concentration of sodium than the surrounding fluid. The neuron also has a higher concentration of potassium inside the cell. The sodium-potassium ion pump is used to maintain the neuron in the resting state. Which of the following statements is true?

a. Remaining at rest requires the use of ATP.
b. Remaining at rest requires an input of sodium.
c. Remaining at rest requires the activation of cotransporters.
d. Remaining at rest requires decreased permeability of the membrane.

Answers

Answer:

a. Remaining at rest requires the use of ATP.

Explanation:

The resting membrane potential is maintained by the sodium-potassium pump. The sodium potassium pump does this by actively pumping sodium ions out of the cell and potassium ions inside the cell in a ratio of 3:2. This movement of ions by the sodium-potassium pump is against their concentration gradient. In a neuron at rest, there are more sodium ions outside the cell than there are inside the cell. Also, there are are more potassium ions inside the cell than there are outside the cell. However, there are ion channels through which these ions enter and leave the cell. Sodium ion channels allow sodium to enter the cell following its concentration gradient, whereas, potassium ion channels allow potassium to leave the cell following its concentration gradient. However, more potassium ions leave the cell than do sodium ions enter the cell because of the higher permeability of the cell to potassium ions.

In order to maintain the resting membrane potential, the sodium potassium pump powered by the hydrolysis of an ATP molecules pumps sodium ions out of the cell and potassium ions into the cell.

Therefore, the correct option is A, as ATP is needed by the sodium-potassium pump in order to maintain the resting membrane potential.

Smallest particle of a substance that can retain its properties & is
composed of one or more atoms *
O A. Atom
O B. Molecules
O C. Matter
O D. Everything

Answers

Answer:

Molecules

Explanation:

2H2 (1) + O2(g) → 2H20 (g)

1. Find the limiting reactant if you start with 30.0 grams of hydrogen and 5.29 grams of oxygen.

2. The actual yield for H2O in the above reaction is 6.72 g, Determine the percent yield for the reaction

when 9.93 grams of hydrogen and excess oxygen react?

Answers

Answer: 1. Oxygen is the the limiting reactant.

2. 7.52%

Explanation:

The balanced chemical equation is:

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]  

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} H_2=\frac{30.0g}{2g/mol}=15.0moles[/tex]

[tex]\text{Moles of} O_2=\frac{5.29g}{32g/mol}=0.165moles[/tex]

According to stoichiometry :

1 mole of [tex]O_2[/tex] require  = 2 moles of [tex]H_2[/tex]

Thus 0.165 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.165=0.331moles[/tex]  of [tex]NH_3[/tex]

Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.

2.  [tex]\text{Moles of} H_2=\frac{9.93g}{2g/mol}=4.96moles[/tex]

[tex]\text{Moles of} H_2O=\frac{6.72g}{18g/mol}=0.373moles[/tex]

As 2 moles of [tex]H_2[/tex] give = 2 moles of [tex]H_2O[/tex]

Thus 4.96 moles of [tex]H_2[/tex] give =[tex]\frac{2}{2}\times 4.96=4.96moles[/tex]  of [tex]H_2O[/tex]

percentage yield = [tex]\frac{\text {actual yield}}{\text {theoretical yield}}=\frac{0.373}{4.96}\times 100=7.52\%[/tex]

Thus the percent yield for the reaction is 7.52%

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