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7nadin3 [17]
3 years ago
11

An example of a pure substance is ____.

Chemistry
1 answer:
Vsevolod [243]3 years ago
6 0

Answer: e) all of these

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Copper

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: water  (H_2O), carbon dioxide (CO_2)

Mixtures are not pure substances as they consist of elements and compounds combined physically and not in any fixed ratio. Example: Air

Thus all given substances are pure substances.

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What is the mass in grams of 9.45*10^24 molecules of methanol (CH3OH)?
Angelina_Jolie [31]
Number of moles:

1 mole ---------- 6.02x10²³ molecules
? moles --------- 9.45x10²⁴ molecules

1 x ( 9.45x10²⁴) / 6.02x10²³ =

9.45x10²⁴ / 6.02x10²³ => 15.69 moles of CH3OH

Therefore:

Molar mass CH3OH = 32.04 g/mol

1 mole ------------ 32.04 g
15.69 moles -----  mass methanol

Mass methanol  = 15.69 x 32.04 / 1 => 502.7076 g


6 0
3 years ago
What volume of a 2.5 M stock solution of acetic acid (HC2H3O2) is required to prepare
Schach [20]

<span> </span>

Answer is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH) = c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH) = 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.

8 0
3 years ago
Complete the following radioactive decay problem.<br>222 86 RN to 4 2 HE + ?​
Andreas93 [3]

Answer:

₈₆²²²Rn    →   ₈₄Po²¹⁸  +  H₂⁴

Explanation:

The given nuclear reaction shows alpha decay.

₈₆²²²Rn    →   ₈₄Po²¹⁸  +  H₂⁴

Properties of alpha radiations:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4  and atomic number less than 2 as compared to parent atom the starting atom.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

₉₂U²³⁸   →   ₉₀Th²³⁴  + ₂He⁴  + energy

8 0
3 years ago
What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)
amid [387]

Answer:

m=59.3gCO_2

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol

Finally, the mass by using the molar mass:

m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2

Best regards.

4 0
3 years ago
An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
Anna [14]

Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

5 0
3 years ago
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