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ozzi
3 years ago
11

Which of the following elements has the highest electronegativity? A. oxygen (O) B. sulfur (S) C. fluorine (F) D. chlorine (Cl)

Chemistry
1 answer:
Katena32 [7]3 years ago
4 0
C. Fluorine with an electronegativity of 4.0
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What volume of 0.4567 m h2so4 is required to neutralize 30.00 ml of 0.3210 m naoh?
sertanlavr [38]
Answer is: volume of H₂SO₄ is 42.1 mL.<span>
Chemical reaction: H</span>₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.<span>
c(H</span>₂SO₄) = 0,4567 M = 0,4567 mol/L.<span>
V(NaOH) = 30 mL </span>÷ 1000 mL/L <span>= 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H</span>₂SO₄) : n(NaOH) = 1 : 2.<span>
n(H</span>₂SO₄) = 0,01926 mol.<span>
V(H</span>₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).<span>
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3 years ago
What law states that energy is never created or destroyed?​
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D. the law of conservation of energy
8 0
3 years ago
True or False: The products of a chemical
attashe74 [19]

Answer:

true

Explanation:

6 0
3 years ago
Read 2 more answers
Student A performed gravimetric analysis for sulfate in her unknown using the same procedures we did . Results of her three tria
Andre45 [30]

Let us first calculate the accepted value of sulfate using the molar mass of sodium sulfate.

The molar mass of sodium sulfate is 2 Na + S + 4 O = 2 (22.99) + 32.07 + 4 (16)

Molar mass of Sodium sulfate = 142.05

Molar mass of sulfate = S + 4 (O) = 32.07 + 4 (16) = 96.07

% of sulfate in sodium sulfate = (Molar mass of SO4 / Molar mass of Na2SO4) x 100

% of sulfate = (96.07/142.05) x 100

% of sulfate = 67.6%

Theoretical value for % Sulfate is 67.6%

Percent error is calculated as  

Percent Error = \left | \frac{Experimental - Theoretical}{Theoretical} \right |

Let us calculate mean for student A.

Mean for student A = (68.6 + 66.2 + 67.1) / 3 = 201.9/3 = 67.3 %

% error for student A = \left | \frac{67.3 - 67.6}{67.6} \right |

% error for student A = 0.44%

Mean for student B = (66.7 +66.6 + 66.5)/3 = 199.8 /3 = 66.6 %

% error for student B = \left | \frac{66.6 - 67.6}{67.6} \right |  

% error for student B = 1.48%

Accuracy is the closeness of experimental values to the true value. This is defined in terms of absolute and percent errors.

Since percent error for student A is lower, we can say that student A was more accurate.

The precision of the data depends on the closeness of experimental values to each other. We can see that experimental values for student B were close to each other.

Therefore we can say that student B was more precise.


8 0
4 years ago
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