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liberstina [14]
3 years ago
13

What advice would you give to a person who handles hydrogen and oxygen in their workplace?I think I have an idea of what advice

to give, but I'm not entirely sure if it's correct. Please help!! I'm really bad at science.
Chemistry
1 answer:
finlep [7]3 years ago
3 0
Regard the principle of utilization of two gas.

Make a consistent control of hardware containing gas.

Make a consistent control of weight diminishing valves giving gas.

No smoking zone.
You might be interested in
What would happen if the lid of a jar filled with helium gad was removed?
yawa3891 [41]
The helium would expand into the air of the room. Depending on how big the room is, and the amount of helium in the jar, it could cause damage to health. The only way this would happen is if the room was small and the jar was big.

I hope this helps!
7 0
2 years ago
19. Which of the following chemical equations represents
Lesechka [4]

Answer:

  • C₃H₈ (g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (l)

(option D. with the proviso that the subscripts of propane's chemical formula must be corrected)

Explanation:

<em>Propane</em> is the saturated hydrocarbon, alkane, with chemical formula C₃H₈ or CH₃CH₂CH₃.

The complete combustion of the hydrocarbons yield carbon dioxide (CO₂) and water (H₂O).

The chemical equation that represents this combustion is:

  • C₃H₈ (g) + O₂(g) → CO₂ (g) + H₂O (l) (skeleton equation: unbalanced)

Once you balance it, you get:

  • C₃H₈ (g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (l)

        Left side           Right side

C         3                          3

H         8                          4×2 = 8

O        5×2 = 10               3×2 + 4 = 10

That equation corresponds to the option D. of the list, with the proviso that the subscripts of propane's chemical formula must be corrected

6 0
3 years ago
The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0
2 years ago
The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) th
seropon [69]

Answer:

(a) a = 5.08x10⁻⁸ cm

(b) r = 179.6 pm  

Explanation:

(a) The lattice parameter "a" can be calculated using the following equation:

\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}      

<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>N_{A}<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>

Hence the lattice parameter is:  

a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}

a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm

(b) We know that the lattice parameter of a FCC structure is:

a = \frac{4r}{\sqrt{2}}

<em>where r: is the atomic radius of Th</em>

Hence, the atomic radius of Th is:

r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm    

I hope it helps you!    

4 0
3 years ago
Read 2 more answers
Which of the following subatomic particles has the LEAST mass?
ivanzaharov [21]
An electron this is the smallest particle and it takes up the least mass because its negative charge makes it harder for electrons to have a place in an atom.
3 0
3 years ago
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