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DIA [1.3K]
2 years ago
14

I need help please???

Chemistry
1 answer:
Andreyy892 years ago
3 0

Answer:

the answer is the 2nd option

Explanation:

distance is divided by rime

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State whether the following statements are true or false (with justification). (a) 1 mol of N2 has more molecules than 1 mol of
MAVERICK [17]

Answer:

A. False.

Every substance contains the same number of molecules i.e 6.02x10^23 molecules

B. False.

Mass conc. = number mole x molar Mass

Mass conc. of 1mole of N2 = 1 x 28 = 28g

Mass conc. of 1mol of Ar = 1 x 40 = 40g

The mass of 1mole of Ar is greater than the mass of 1mole of N2

C. False.

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of Ar = 40g/mol

The molar mass of Ar is greater than that of N2.

Explanation:

7 0
2 years ago
Significant figures are an indicator of the certainty in measurements true or false?
IRISSAK [1]
True. 

SF are used for simplifying figures in a measurement to produce a more accurate reading. 
3 0
3 years ago
Describe what an alchemist does
Zielflug [23.3K]
Yes what he saiddddd
6 0
2 years ago
in the following reaction, how many grams of benzene (C6H6) will produce 42 grams of CO2? 2C6H6 + 15O2 → 12CO2 + 6H2O
Mrrafil [7]

Answer: -

12.41 g

Explanation: -

Mass of CO₂ = 42 g

Molar mass of CO₂ = 12 x 1 + 16 x 2 = 44 g / mol

Number of moles of CO₂ = \frac{42}{44 g/mol}

= 0.9545 mol

The balanced chemical equation for this process is

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

From the balanced chemical equation we see

12 mol of CO₂ is produced from 2 mol of C₆H₆

0.9545 mol of CO₂ is produced from \frac{2 mol C6H6 x 0.9545 mol  CO2}{12 mol CO2}

= 0.159 mol of C₆H₆

Molar mass of C₆H₆ = 12 x 6 + 1 x 6 =78 g /mol

Mass of C₆H₆ =Molar mass x Number of moles

= 78 g / mol x 0.159 mol

= 12.41 g

8 0
3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
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