Why do we test for free chlorine in drinking water?

Answers

Answer 1

Answer:

The presence of free chlorine in drinking water is correlated with the absence of most disease-causing organisms, and thus is a measure of the potability of water


Related Questions

1.20×10−8s to nanoseconds

Answers

Answer:

There are 12 nanoseconds in [tex]1.2\times 10^{-8}\ s[/tex].

Explanation:

We need to convert [tex]1.2\times 10^{-8}\ s[/tex] to nanoseconds.

We know that,

[tex]1\ s=10^9\ ns[/tex]

Now using unitary method to solve it such that,

[tex]1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns[/tex]

So, there are 12 nanoseconds in [tex]1.2\times 10^{-8}\ s[/tex].

What was the purpose of letting the transformed cells sit in LB for a few minutes before spreading them onto the plates?

A. This allows time for the cells to express the antibiotic resistance gene

B. This allows the cells to take up the plasmid after the heat shock procedure

C. This allows time for the cells to warm up before plating

D. This allows cells time to start glowing green

Answers

I think c ... C this is the correct

Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.

DATA
Copper
Lead
Mass of Metal




Initial Temperature of Metal




Mass of Water




Initial Temperature of Water




Final Temperature of Water






CALCULATIONS
Copper
Lead
Temperature Change of Water




Heat Gained by Water




Heat Lost by Metal




Temperature Change of Metal




Specific Heat of Metal




% Error






Use the Gizmo to mix 200 g of lead at 100 °C with 1,000 g of water at 20 °C. Record the data and answers to the calculations for lead in the 2 tables above.

Metal
Accepted Specific Heat (J/g⁰C)
aluminum
0.900
lead
0.160
copper
0.385
tin
0.228
steel
0.460


Insert a picture of your work for the calculations table here.

Answers

Answer:

Explanation:Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.

DATA

Copper

Lead

Mass of Metal

A 38.22 mL aliquot of weak acid that has a concentration of 0.882 M will be titrated with 0.289 M NaOH. Calculate the pH of of the solution upon neutralization of half of the weak acid. The Ka of the acid is 6.8×10-7.

Answers

Answer:

pH = 6.167

Explanation:

The weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H₂O

Where 1 mole of HX with 1 mole of NaOH produce 1 mole of NaX (The conjugate base of the weak acid).

Now, using H-H equation:

pH = pKa + log [NaX] / [HX]

Where pH is the pH of the buffer

pKa is -log Ka = 6.167

And [NaX] [HX] are the molar concentrations of each specie

Now, at the neutralization of the half of HX, the other half is as NaX, that means:

[NaX] = [HX]

And:

pH = pKa + log [NaX] / [HX]

pH = 6.167 + log 1

pH = 6.167

Glucose and alcohol Contain hydrogen
but they are not adds Why?​

Answers

Answer:

Ans. Alcohols and glucose though contain hydrogen but do not ionise in the solution to produce H+ ions. This is proved by the fact that there solutions do not conduct electricity.

In what direction does thermal energy transfer?

Answers

Answer:

Either direction

Explanation:

If you cook food it would go from cold to hot. If you were to put a drink in the fridge it would go from hot to cold. So it is either direction.

Which of these would a mechanical engineer do?

Answers

Answer:

I dont see the answers

Explanation:

How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?

Answers

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead =  177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g       (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead =[tex]\frac{177.7}{207.2}\times 1310=1123kJ=1123000J[/tex]

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Need help is label a warm area or is b Orr are they both warm

Answers

Answer:

your image is a blavk screen so we cant tell

Explanation:

What's a household item that has  electrical energy

Answers

Answer:

Washing machine.

Dryer.

Television.

Cell phone.

Explanation:

Answer:

Washer and dryer, TV, Cell phone

Explanation:

Use mass-mass calculation, determine how many grams of sodium chloride SHOULD have been produced from the amount of sodium bicarbonate in mass (B.)

Answers

Answer:

where is the question???

Which of the following would be considered renewable resources? You may choose more than one.


oil

corn

clean air

forests

Answers

Answer:

clean air

Explanation:

someone already answer but clean air

A chemist needs to determine the concentration of a solution of nitric acid, HNO3. She puts 905 mL of the acid in a flask along with a few drops of indicator. She then slowly adds 0.200 M Ba(OH)2 to the flask until the solution turns pink, indicating the equivalence point of the titration. She notes that 235 mL of Ba(OH)2 was needed to reach the equivalence point.

Required:
a. How many moles of Ba(OH)2 are present in 225 mL of 0.200 M Ba(OH)2?
b. How many moles of HNO3 are present if 4.50 x 10^2 mol of Ba(OH)2 was needed to neutralize the acid solution? (Express your answer numerically in moles)

Answers

Answer:

See explanation

Explanation:

The equation of the reaction is;

2HNO3(aq) + Ba(OH)2(aq) -------->Ba(NO3)2(aq) + 2H2O(l)

a) The number of moles of  Ba(OH)2  = 225/1000 *  0.200

= 0.045 moles

From the reaction equation;

2 moles of HNO3 require 1 mole of  Ba(OH)2

x moles of HNO3 require 4.50 x 10^2 mol of Ba(OH)2

x = 4.50 x 10^2 * 2/1

x = 9  x 10^2 moles of HNO3

Is lighting a match a chemical change?

Answers

Answer:

yes

Explanation:

Lighting a match and letting is burn is an example of a chemical change.

Matches use sulfur, phosphate and a friction agent held together by a binding agent. Together, the oxygen and sulfur burn slowly, igniting the wood of the match for a flame that lasts long enough to see by, light a candle or ignite a campfire.

What causes the lines in the spectrum for elements

Answers

They have to connect atoms to cause a chemical reaction

Is a cold and b is warm? Please help

Answers

Answer:

Label A shows cold area on the mug, while label B shows a warm area.

Explanation:

Have a good day

A reaction was experimentally determined to follow the rate law, Rate = k[A] where k = 0.15 s-1. Starting with [A]o = 0.225M, how many seconds will it take for [A]t = 0.0350M?

Answers

Answer:

The time the reaction takes is 12.4s

Explanation:

A reaction that follows the rate law:

Rate =k[A] is order 1 and follows the equation:

Ln[A] = -kt + ln[A]₀

Where [A] is concentration of the reaction after time t: 0.0350M

k is rate constant = 0.15s⁻¹

t is time in seconds

[A]⁰ is initial concentration = 0.225M

Ln[0.0350] = -0.15s⁻¹*t + ln[0.225]

-1.86075 =  -0.15s⁻¹*t

12.4s = t

The time the reaction takes is 12.4s

BIOCR
1 Co
> Co

Answer the question below. Use the rubric in the materials for help if needed.
>
Describe the process used to determine how many atoms of each element are in 2Ca3(PO4)2. Show all
your work.

Answers

Answer:

Six atoms of calcium, four of phosphorous and sixteen of oxygen for a total of twenty six

Explanation:

Hello there!

In this case, according to the molecular formula of the two moles of calcium phosphate:

[tex]2Ca_3(PO_4)_2[/tex]

Thus, in order to calculate the atoms of each atom, it is necessary to multiply the two in front of the formula by the subscripts in the reaction:

[tex]atoms Ca=2*3=6\\\\atoms P=2*2=4\\\\atoms O=2*2*4=16[/tex]

Thus, we obtain six atoms of calcium, four of phosphorous and sixteen of oxygen for a total of twenty six.

Best regards!

Which compound contains three elements?
A. Aluminium chloride
B. iron(III) oxide
C. potassium oxide
D. sodium carbonate

Answers

Answer:

Sodium Carbonate

Explanation:

Sodium carbonate is made of three elements, that are, sodium, carbon and oxygen.

Its formula is Na2CO3.

Hope it helps:)

The half-life of argon-39 is 269 years. It decays into krypton-39. After 1,076 years, what fraction of the original amount of argon-
1/16
1/4
1/2
1/8

Answers

Answer:

The correct answer is - 1/16.

Explanation:

The half-life is the time that is required to decay nuclei to the half amount of its original amount in a radioactive sample. It is represented by the t1/2. The time that is required to reduce to half of its initial value is known as the half-life of a particular sample.

Half-life of argon-39 = 269 years

total time is taken in complete decay = 1076 years

Number of half-lives required = 1076/269

= 4

So the original amount of the sample = (1/2)^n

n = number of half-life

= 1/2^4

= 1/16

The correct answer is = 1/16

How many moles are in 6 x 10^23 molecules of H2O

Answers

One mole represents 6.022∙1023 separate entities, just like one dozen represents 12 objects. So, if there are 6.022∙1023 H2O molecules, that is the same as one mole of water.

A gas cylinder is filled with 5.50 moles of oxygen gas at 83°C. The piston is compressed to yield a pressure of
400.0 kPa. What is the volume inside the cylinder?

Answers

Answer:

volume=0.04322m3

Explanation:

acording to ideal gas equation that PV=nRT

2. A 0.45g sample of mineral ore is analyzed for chromium and found to contain
0.560mg Cr2O3. Express the concentration of Cr2O3 in the sample as:
(a) Percent (%)
(b) Parts per thousand (ppt)
(c) Parts per million (ppm)
(d) Parts per billion (ppb)​

Answers

Answer:

Explanation: I don't know

What can you infer about the air pressures over the land and ocean? Explain your answer.


Answers

Answer:

Answer: During the day, the sun heats up both the ocean surface and the land. ... The wind will blow from the higher pressure over the water to lower pressure over the land causing the sea breeze. The sea breeze strength will vary depending on the temperature difference between the land and the ocean.

Explanation:

suppose you are analyzing an antacid that contains Mg(OH)2 and Al(OH)3 in a 1:1 molar ratio. The antacid was neutralized with an excess of 0.500 M HCl, and the excess HCl was then back titrated with 0.500 M NaOH. The following data was obtained: Volume of 0.500 M HCl utilized: 38.25 mL Volume of 0.500 M NaOH utilized: 3.60 mL Calculate the masses of Mg(OH)2 and Al(OH)3 contained in the antacid.

Answers

Answer:

0.202g of Mg(OH)₂ and 0.270g of Al(OH)₃

Explanation:

The OH⁻ ions of the Mg(OH)2 and Al(OH)3 reacts with the HCl. The HCl in excess reacts with the NaOH. To solve this question we must find the moles of HCl that react in the beginning with the antiacid and as the ratio is 1:1 we can find the moles of Mg(OH)2 and Al(OH)3. The reactions are:

OH⁻ + HCl → Cl⁻ + H₂O

HCl + NaOH → NaCl + H₂O

Moles HCl added:

0.03825L * (0.500mol / L) = 0.019125 moles

Moles NaOH utilized:

0.0036L * (0.500mol / L) = 0.0018 moles

Moles HCl that react = Moles OH⁻ in the antiacid

0.019125 moles - 0.0018moles =

0.017325 moles OH⁻ in the antiacid

As the ratio is 1:1. the moles Mg(OH)2 = Moles Al(OH)3

And these 5 hydroxides produce 0.017325 moles:

5X = 0.017325 moles

X = 3.465x10⁻³ moles

Where X are moles of both Mg(OH)2 and Moles Al(OH)3

Mass -Molar mass: 58.3197 g/mol-:

3.465x10⁻³ moles * (58.3197g/mol) =

0.202g of Mg(OH)₂

And

Mass -Molar mass: 78g/mol-:

3.465x10⁻³ moles * (78g/mol) =

0.270g of Al(OH)₃

The air temperature in a sealed, insulated box.. 20°C An ice cube at O'C is
placed in the box where it slowly melts. How and why does the melting of the ice
affect the air temperature in the box?

Melting is endothermic so the air temperature increases

Melting is exothermic, so the air temperature decreases


Melting is endothermic, so the air temperature decreases

Melting is exothermic, so the air temperature increases

Answers

Answer:

Melting is endothermic, so the air temperature decreases.

Explanation:

Hello there!

In this case, according to the given statement, it is possible to infer that the ice is melt because energy is applied to the ice and thereafter its temperature increases; this is possible because the joined particles of a solid substance need energy to undergo such a separation that they become more far away to each other and therefore transcend to the liquid phase due to the new molecules arrangement. Thus, the answer is Melting is endothermic, so the air temperature decreases because as the ice heats up, the air cools down as it gives it energy to the ice.

Best regards!

What is the highest occupied level of Carbon

Answers

Answer:

Highest occupied level of carbon is 3

Explanation:

its called Alkynes

Explanation:

4

trust me!

#carryonlearning❤

I need help pls!!!!

Answers

Answer:

A

Explanation:

jsgsnsbnss care about explaimi

Consider the reaction, C2H4(g) + H2(g) - C2H6(8), where AH = -137 kJ. How many kilojoules are released when 3.5 mol of CH4
reacts?
480 kJ are released
20 x 103 kJ are released
570 kJ are released
137 kJ are released​

Answers

Answer: 480 kJ of energy is released when 3.5 mol of [tex]C_2H_4[/tex] reacts.

Explanation:

The balanced chemical reaction is:

[tex]C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)[/tex]  [tex]\Delta H=-137kJ[/tex]

Thus it is given that the reaction is exothermic (heat energy is released) as enthalpy change for the reaction is negative.

1 mole of [tex]C_2H_4[/tex] on reacting gives = 137 kJ of energy

Thus 3.5 moles  of [tex]C_2H_4[/tex] on reacting gives = [tex]\frac{137}{1}\times 3.5=480 kJ[/tex] of energy

Thus 480 kJ of energy is released when 3.5 mol of [tex]C_2H_4[/tex] reacts.

I will mark brainliest plz help me!!!
What kind of weather forms with an occluded front?

Answers

Occluded fronts usually form around areas of low atmospheric pressure.
There is often precipitation along with the occluded front with clouds.
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