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Umnica [9.8K]
3 years ago
15

What is the correct prefix for 5? A. Deca- B. Penta- C. Tetra- D. Hexa-

Chemistry
2 answers:
Art [367]3 years ago
6 0

Answer:

Its B penta

Explanation:

a_sh-v [17]3 years ago
5 0

Answer:

B. penta

you’re welcome

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What is the second step in the classification of living things?
Leya [2.2K]
I'll go with B, have a good day
6 0
3 years ago
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What kind of electrons are involved in chemical bonding?
mamaluj [8]
Bonding electrons are involved in chemical bonding these electrons have their valnce shell incomplete
6 0
3 years ago
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions. If 0.212 g of KHP are dissolve
pochemuha

0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.

The balanced neutralization equation is:

NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)

  • Step 1: Calculate the reacting moles of KHP.

0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.

0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol

  • Step 2: Determine the reacting moles of NaOH.

The molar ratio of NaOH to KHP is 1:1.

1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH

  • Step 3: Calculate the molarity of NaOH.

1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.

[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M

0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Learn more about titration here: brainly.com/question/4225093

3 0
2 years ago
To test the purity of sodium bicarbonate, you dissolve a 3.50g sample in water and add sulfuric acid. if 1.04g of carbon dioxide
Andrews [41]

Answer is: the percent purity of the sodium bicarbonate is 56.83 %.

1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.

2. m(NaHCO₃) = 3.50 g

n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).

n(NaHCO₃) = 3.50 g ÷ 84 g/mol.

n(NaHCO₃) = 0.042 mol.

3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.

n(CO₂) = 0.042 mol.

m(CO₂) = 0.042 mol · 44 g/mol.

m(CO₂) = 1.83 g.

4. the percent purity = 1.04 g/1.83 g  ·100%.

the percent purity = 56.8 %.

8 0
3 years ago
How many grams of F are in 12.56 g of SF6? h.
natali 33 [55]

Answer:

9.80 g

Explanation:

The molecular mass of the atoms mentioned in the question is as follows -

S = 32 g / mol

F = 19 g / mol

The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol

The mass of 6 F = 6 * 19 = 114 g /mol .

The percentage of F in the compound =

mass of 6 F / total mass of the compound * 100

Hence ,  

The percentage of F in the compound = 114 g /mol  / 146 g / mol * 100

78.08 %

Hence , from the question ,

In 12.56 g of the compound ,

The grams of F = 0.7808 * 12.56 = 9.80 g

4 0
3 years ago
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