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alisha [4.7K]
3 years ago
5

SiO2 + 4HF = SiF4 + 2H2O, what mass of HF is necessary to produce 12.91g of the SiF4?

Chemistry
1 answer:
ahrayia [7]3 years ago
7 0
i dont know broooooooooooooooooooooooooooooooooooooooooooioooooooooooooooooooooooooooooooooo i wish i knew
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An atom x contains 14 protons, 13 electrons, and 12 neutrons, and atom y contains 14 protons, 14 electrons, and 12 neutrons. wha
Vika [28.1K]
<span>a. x and y are atoms of the same element. 

If both atoms contain the same amount of protons, they are always the same element.
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7 0
3 years ago
Read 2 more answers
Vamos para playa oh te rajas lol
Anuta_ua [19.1K]

Answer:

let's go to the beach or u will relax lol

6 0
3 years ago
1]
balu736 [363]

Explanation:

Given -

  • An organic compound gives H₂ gas with Na
  • On treatment with alkaline iodine it gives yellow ppt.
  • On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)

To Find -

  • Name the compound and write the reaction involved

Now,

Let A be the organic compound.

Then,

  1. A + Na → + H₂↑
  2. A + I₂ → CHI₃ (yellow ppt.)
  3. A + CrO₃ + H⁺ → C₂H₄O

Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

  • Functional group of aldehyde = —CHO

And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

It means,

We know that 1° alcohol on oxidation gives aldehyde.

Here it gives 2 Carbon aldehyde.

It means,

Here 2 Carbon and 1° alcohol is used.

Now,

Its cleared that Compound A is Ethanol.

Reaction Involved -

  1. CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
  2. CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
  3. CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO

6 0
3 years ago
9.44 x 10^3 In standard form
Marta_Voda [28]
9440 is standard form.
5 0
3 years ago
At its critical point, ammonia has a density of 0.235 g cm23. You have a special thick­walled glass tube that has a 10.0­mm outs
salantis [7]

Answer:

\large \boxed{\text{69.3 mg}}

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}

4 0
3 years ago
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