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Misha Larkins [42]
3 years ago
9

Choose the number of significant figures 60.0

Chemistry
2 answers:
GenaCL600 [577]3 years ago
8 0

Answer:

3 significant figures

Explanation:

The rules for deciding which digits in a measurement are significant are as follows:

1) All nonzero digits are significant.

2) Sandwiched (or embedded) zeros, those between significant digits, are significant.

3) Leading zeros, which are zeros at the beginning of a decimal number less than 1, are not significant.

4) Trailing zeros, which are zeros at the end of a number, are significant only if the number has a decimal point.

By rule 4 since 60.0 has a decimal point all 3 figures are significant

oksian1 [2.3K]3 years ago
4 0

 

In this case there are 3 significant figures.

Any zeroes after another number, (i.e. 9300.000), you consider these numbers.  

If there is something like 0.0060, this would be two significant digits.  

Zeroes before a number are not significant, anything after are.  

Hope this helps any confusion!


Please mark me brainliest!

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In 1986 an electrical power plant in taylorsville, georgia, burned 8,376,726 tons of coal, a national record at that time. assum
RideAnS [48]
<span>C + O2 → CO2 (8,376,726 tons) x (0.80) / (12.01078 g C/mol) x (1 mol CO2/ 1 mol C) x (44.00964 g CO2/mol) = 24,555,054 tons CO2</span>
6 0
3 years ago
Read 2 more answers
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
2 years ago
Calculate the mass fraction of sodium chloride in the solution if 20 g of it is dissolved in 300 ml of water.
MA_775_DIABLO [31]

The mass fraction of sodium chloride is 0.0625

<h3>What is the mass fraction of sodium chloride in the solution?</h3>

The mass fraction of sodium chloride is the ratio of the mass of sodium chloride to the total mass of the solution.

The mass fraction of sodium chloride is determined as follows;

mass of sodium chloride = 20 g

  • mass of water = volume * density

density of water = 1 g/mL

volume of water = 300 mL

mass of water = 300 mL * 1 g/mL

mass of water = 300 g

total mass of solution = 20 + 300 = 320 g

mass fraction of sodium chloride = 20/320

mass fraction of sodium chloride = 0.0625

Learn more about mass fraction at: brainly.com/question/14783710

#SPJ1

5 0
1 year 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
How many atoms are in oxygen
Aleonysh [2.5K]
There are only 2 atoms in an Oxygen molecule
4 0
3 years ago
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