Help! Can someone please explain and break down the answers to both? I’m so confused!1) The moles of H2O that f am be obtained from 15.0mL of 0.250 M HCl

2) the volume of 0.150 M KMnO4 needed to replaced 1.85 mol MnCl2

Answers

Answer 1

Answer: Hope this helps you!!

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A 0.2264 g sample of a pure carbonate, XnCO3(s) , was dissolved in 50.0 mL of 0.1800 M HCl(aq) . The excess HCl(aq) was back titrated with 24.90 mL of 0.0980 M NaOH(aq) . How many moles of HCl react with the carbonate
A system absorbs 12 J of heat from the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the internal energy ΔEth of the system?
Research Hypatia's achievements in the worldof science. What is she most known for?Write down three interesting facts about herlife.
Which of the following chemical equations violates the law of conservation of mass? (1 point) Li2SO4 + 2Ca(NO3)2 yields 2LiNO3 + 2CaSO4 Ti + 2Cl2 yields TiCl4 2C2H2 + 5O2 yields 4CO2 + 2H2O MgSO4 + 2NaOH yields Mg(OH)2 + Na2SO4

A car travels for 0.5 hours and goes 25 miles. What is the car's speed? (Remember sad + t)

Answers

Answer:

s = 22.352 m/s

Explanation:

Given data:

Time taken = 0.5 hours

Distance cover = 25 miles

Car speed = ?

Solution:

First of all we will convert the units.

1 hour = 3600 sec

0.5 hr ×3600 sec / 1hr = 1800 sec

miles to meter:

25 mi × 1609 m/ 1 mi = 40233.6 m

Formula:

s = d/t

s = speed

d = distance

t = time

Now we will put the values in formula.

s = 40233.6 m / 1800 sec

s = 22.352 m/s

Final answer:

The speed of the car, calculated by dividing the distance travelled (25 miles) by the time taken (0.5 hours), is 50 miles per hour.

Explanation:

The subject of the question falls under Mathematics, specifically a section of it named rate, time, and distance problems. The problem is asking us to calculate the speed of a car which can be obtained by dividing the distance travelled by the time taken. Given that the distance travelled by the car is 25 miles and the time taken is 0.5 hours, we can calculate the speed as follows.

Denote the Speed as S, the distance as D, and the time as T.
Since Speed = Distance/Time, we substitute our values into this equation
Therefore, Speed = 25 miles / 0.5 hour = 50 miles/hour

So, the speed of the car is 50 miles per hour.

Learn more about Speed Calculation here:

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Determine the concentrations of K2SO4, K+, and SO42− in a solution prepared by dissolving 2.07 × 10−4 g K2SO4 in 2.50 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of SO42−. Ignore any reactions with water.

Answers

Answer:

[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ; [SO4⁻²] = 4,75x10⁻⁷M

SO4⁻²: 0.045ppm ; K⁺: 0.037ppm

[SO4⁻²] = 4,70x10⁻⁷ F

Explanation:

Determine the equation

K2SO4 → 2K⁺ + SO4⁻²

Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion

Molar mass K2SO4: 174.26 g/m

Moles of K2SO4: grams / molar mass

2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles

Molarity: Moles of solute in 1 L of solution

1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)

K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M

SO4⁻²: 4,75x10⁻⁷ M

1 mol of K2SO4 has 2 moles of K and 1 mol of SO4

1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.

1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)

2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)

These grams are in 2.5 L of water, so we need μg/mL to get ppm

2.5 L = 2500 mL

1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)

9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)

113.35 μg /2500 mL = 0.045ppm

92.6 μg /2500 mL = 0.037ppm

Formal concentration of SO4⁻² :

Formality = Number of formula weight of solute / Volume of solution (L)

(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F

I have always enjoyed eating tuna fish. Unfortunately,a study of the mercury content of canned tuna in 2010 foundthat chunk white tuna contains 0.6 ppm Hg and chunk lighttuna contains 0.14 ppm. (S. L. Gerstenberger, A. Martinson,and J. L. Kramer, Environ. Toxicol. Chem. 2010, 29, 237.) TheU.S. Environmental Protection Agency recommends no morethan 0.1 mg Hg/kg body weight per day. I weigh 68 kg. Howoften may I eat a can containing 6 ounces (1 lb 5 16 oz) ofchunk white tuna so that I do not average more than 0.1 mgHg/kg body weight per day? If I switch to chunk light tuna,how often may I eat one can?

Answers

Answer:

You can eat one chunk white tuna of 6 oz every 21,5 minutes

And, you can eat one chunk linght tuna of 6 oz every 5 minutes

Explanation:

The exposure to mercury may cause serious health problems, and is a threat to the development of the child in utero and early in life.

If you weight 68 kg you can eat:

68 kg * 0,1mgHg/ kg = 6,8 mg Hg per day

Chunk white tuna contains 0,6 mg Hg/kg

Thus, you can eat:

6,8 mg Hg * 1 kg tuna/ 0,6 mg Hg = 11,33 kg of chunk white tuna per day

In ounces:

11,33 kg * 35,274oz/ 1 kg = 400 oz per day

You can eat 66,7 6 ounces of chunk white tuna per day. One every 21,5 minutes

Chunk light tuna contains 0,14 mg Hg/kg

Thus, you can eat:

6,8 mg Hg * 1 kg tuna/ 0,14 mg Hg = 48,57 kg of chunk white tuna per day

In ounces:

48,57 kg * 35,274oz/ 1 kg = 1713 oz per day

You can eat 285,6 6 ounces of chunk white tuna per day. One every 5 minutes

I hope it helps!

Which three are advantages of asexual reproduction?A:Offspring are more likely to survive environmental changes.

B:Some offspring are more likely to survive a disease.

C:Less energy is required to reproduce.

D:The population can increase from only one parent.

E:The population can increase quickly.

Answers

C, D, and E

A and B cannot be true because asexual reproduction means the parent organism is essentially creating clones of itself, providing no variation in DNA and making all offspring vulnerable to the same environmental changes and diseases as the parent.

Answer:

sorry if I get this wrong I think it is C

Explanation:

a length of #8 copper wire (radius = 1.63 mm) has a mass of 24.0 kg and a resistance of 2.061 ohm per km. what is the overall resistance of the wire

Answers

m = pVol

p : copper density

Vol: volume

Vol = Al

A: area

L: length

m = p * A * L

L = m/p*A

To find the total resistance

Rtotal = R per Km * L/1000

Actually I think the question is missing the copper density.

The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process with a rate constant of 6.5 1/hours. What is the half-life in hours of the breakdown reaction? Enter to 4 decimal places.

Answers

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

The half-life of the breakdown reaction is 0.1066 h

The half-life of a substance is simply defined as the time taken for half of the original substance to decay.

The half-life of a first order reaction can be obtained by the following equation:

$t_(1/2) = (0.693)/(K)$