1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
FromTheMoon [43]
2 years ago
13

(pls help me)

Chemistry
2 answers:
kari74 [83]2 years ago
7 0
Claim 1 Sounds more convincing then claim 2
KATRIN_1 [288]2 years ago
5 0

Answer:

Claim 1

Explanation:

You might be interested in
A waste treatment settling tank treats an industrial waste inflow (QIN) of 0.2 m3/s with a suspended particulate concentration (
iren2701 [21]

The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.

<h3>How to calculate the effluent flow?</h3>

It should be noted that the total inflow will be equal to the total outflow. Therefore,

0.2 + 0.048 = 0.05 + We

Collect like terms

Qe = 0.2 + 0.048 - 0.05

Qe = 0.198m³/sec

The concentration will be:

= (360 × 1000)/0.05

= 7200mg/L.

Learn more about effluent flow on:

brainly.com/question/22714269

#SPJ1

8 0
2 years ago
One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

n(\mathrm{O_2}) = n(\mathrm{H}_2).

Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

m = n \cdot M.

In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
  • Right-hand side: \displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9.

Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0
3 years ago
Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0
3 years ago
Please identify each element associated with the orbital notation or electron configuration. Note: One “slash” is one electron,
mylen [45]

Picture is upside

Explanation:

I can't read upsidedown

4 0
2 years ago
Read 2 more answers
1. How many protons does 14/6 C contain? What element is this?
Harman [31]

Answer:

It has 6 protons and its Carbon 14

Explanation:

3 0
3 years ago
Other questions:
  • Explain in terms of both electrons and energy how the bright line spectrum of an element is produced
    9·1 answer
  • A chemical that can superheat or give off poisonous vapors when it comes in contact with air or water is called
    8·1 answer
  • What drug is the greatest challenge for Forensic Science? What drug in specific is a threat in Forensics.
    10·2 answers
  • What properties of a substance determine how that substance will react when combined with other substances?
    8·2 answers
  • Which is NOT a type of fracture?
    14·1 answer
  • Decide which element probably has a boiling point most and least similar to the boiling point of potassium.
    8·1 answer
  • How many molecules of Helium are in 1.0 mole?
    6·1 answer
  • *CHEMISTRY*
    5·2 answers
  • Help plz:)))I’ll mark u Brainliest
    12·1 answer
  • A scientist has two samples of a substance. Both samples have the same temperature. One sample has a mass of 10 g. The other sam
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!