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dedylja [7]
3 years ago
5

Does SiO2 also equal Si2O4? (silicon atom 1 oxygen atom 2 = silicon atom 2 oxygen atom 4?)

Chemistry
1 answer:
Norma-Jean [14]3 years ago
8 0

Answer:

7

Explanation:

that is the anwer of the question

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A cylinder of argon gas contains 50.0 L or Ar at 18.4 atm and 'C. How many moles of argon are in the cylinder?
makkiz [27]

<em>A cylinder of argon gas contains 50.0 L of  Ar  at 18.4 atm and 127 °C. How many moles of argon are in the cylinder?</em>

<em />

Number moles of Argon : 28.03

<h3>Further explanation</h3>

Given

The volume of gas=50 L

P = 18.4 atm

T = 127+273=400 K

Required

moles of Argon

Solution

Use ideal gas Law :

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{18.4\times 50}{0.08205\times 400}\\\\n=28.03

5 0
3 years ago
Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2
Elena-2011 [213]

Answer:

Approximately 0.224\;\rm L, assuming that this reaction took place under standard temperature and pressure, and that \rm CO_2 behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that \rm CaCO_3 is the limiting reactant. How many moles of \rm CO_2 would be produced?

Look up the relative atomic mass of \rm Ca, \rm C, and \rm O on a modern periodic table:

  • \rm Ca: 40.078.
  • \rm C: 12.011.
  • \rm O: 15.999.

Calculate the formula mass of \rm CaCO_3:

\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the number of moles of formula units in 1\; \rm g of \rm CaCO_3 using its formula mass:

\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}.

In the balanced chemical equation, the ratio between the coefficient of \rm CaCO_3 and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1.

In other words, for each mole of \rm CaCO_3 formula units consumed, one mole of \rm CO_2 would be produced.

If \rm CaCO_3 is indeed the limiting reactant, all that approximately 1.00\times 10^{-2}\; \rm mol of \rm CaCO_3\! formula would be consumed. That would produce approximately 1.00\times 10^{-2}\; \rm mol\! of \rm CO_2.

On the other hand, assume that \rm HCl is the limiting reactant.

Convert the volume of \rm HCl to \rm dm^{3} (so as to match the unit of concentration.)

\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 5.00\times 10^{-2}\; \rm dm^{3} of this \rm 0.05\; \rm mol \cdot dm^{-3}

\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}.

Notice that in the balanced chemical reaction, the ratio between the coefficient of \rm HCl and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}.

In other words, each mole of \rm HCl molecules consumed would produce only 0.5\;\rm mol of \rm CO_2 molecules.

Therefore, if \rm HCl is the limiting reactant, that 2.50 \times 10^{-3}\; \rm mol of \rm HCl\! molecules would produce only one-half as many (that is, 1.25\times 10^{-3}\; \rm mol) of \rm CO_2 molecules.

If \rm CaCO_3 is the limiting reactant, \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. However, if \rm HCl is the limiting reactant, 1.25\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be produced.

In reality, no more than \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. The reason is that all \rm CaCO_3 would have been consumed before \rm HCl was.

After finding the limiting reactant, approximate the volume of the \rm CO_2\! produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately 22.4\; \rm L.

If \rm CO_2 behaves like an ideal gas, the volume of that \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be approximately \rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L.

3 0
3 years ago
Which of these periods contain elements with electrons in s, p, d, and f orbitals?
zmey [24]
I think the answer is D. Periods 6-7, Because t<span>he periods that would contain elements with electrons in s, p, d, and f orbitals are periods 6-7. 

(I hope this helps you a lot! :) Have a nice day!)</span>
7 0
3 years ago
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
Murrr4er [49]

Answer:

Explanation:

A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters

PV =nRT

at STP  P= 1atm. T= 273 K

n is the number of moles.  O2 has a molar mass of 32.

30 gm of O2 is 30/32= 0.94 =n

PV = nRT

at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K

SO

1 X V = 0.94 X 0.082 X 273

using high school freshman algebra,

V= 0.94 X 0.082 X 273 = 21L

using high school algebra I,

V=

6 0
2 years ago
The value of a scientific variable:
beks73 [17]

Answer:

A

Explanation:

4 0
3 years ago
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