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Alenkasestr [34]
3 years ago
5

A ____________ is a chemical solution that keeps ph within normal limits by absorbing excess hydrogen, or h+, ions and hydroxide

, or oh-, ions.
Chemistry
2 answers:
garri49 [273]3 years ago
8 0

Answer:

Buffer.

Explanation:

A buffer solution is a solution made out of a weak acid/base and its conjulgated base/acid. This kind of solution is special because it is resistent so pH change because it can react with acids and with bases, thus it keep pH withing a determined range as long as it is not fully consumed by excess adition of acid/base.

TiliK225 [7]3 years ago
5 0
its a buffer solution
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The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mnenie [13.5K]

Answer:

3.1 kg

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O

Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.

The molar mass of C₈H₁₈ is 114.23 g/mol.

1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol

Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈

The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.

Step 4: Calculate the mass corresponding to 70 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

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2 years ago
suppose the burner under the pan of boiling water is turned to a higher settings. how will this affect the temperature of the wa
andrezito [222]

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Explanation:

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This chart lists four elements from the periodic table.
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A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027
irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

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The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

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C = 1.741x10⁻⁵mol / 0.025 L

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