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mr Goodwill [35]
3 years ago
6

Please help ASAP?!?!

Chemistry
1 answer:
Leto [7]3 years ago
7 0

Question: What do you need help with?

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A) What volume of base was needed to reach the equivalence point?
Elan Coil [88]

Answer:

Equivalence point titration

Explanation: Equivalent  point: point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water.

8 0
3 years ago
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Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c
garri49 [273]
The reaction for the combustion of methane can be expressed as follows.
                           CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
                            moles of CO2 = (5.6 L) / (22.4 L/1 mole)
                             moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
                              moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
                                                    = 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane. 
5 0
4 years ago
Read 2 more answers
A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine if each addition would exceed the capacity of the b
likoan [24]

Answer:

no one additions exceed the capacity of the buffer

Explanation:

given

Volume buffer = 500.0 mL = 0.5 L

mol HNO₂ = 0.5 L × 0.100 mol/L = 0.05 mol HNO₂

mol NO₂⁻ = 0.5 L × 0.150 mol/L = 0.075 mol NO₂⁻

solution

we know when any base more than 0.05 (HNO2) than exceed buffer capacity

and when any base more than 0.075 (KNO2) than exceed buffer capacity

when we add 250 mg NaOH (0.250 g)

than molar mass NaOH =40 g/mol

and mol NaOH = 0.250 g ÷ 40g/mol

mol NaOH  = 0.00625 mol

0.00625 mol NaOH will be neutralized by 0.00625 mol HNO₂

so it would not exceed the capacity of the buffer.

and

when we add 350 mg KOH (0.350 g)

than molar mass KOH =56.10 g

and mol KOH = 0.350 g ÷ 56.10 g/mol

mol KOH = 0.0062 mol

here also capacity of the buffer will not be exceeded

and

now we  add 1.25 g HBr

than molar mass HBr = 80.91 g/mol

and mol HBr = 1.25 g  ÷ 80.91 g/mol

mol HBr = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻  

so the capacity will not be exceeded.

and

we add 1.35 g HI  

molar mass HI = 127.91 g/mol

so mol HI = 1.35 g ÷ 127.91 g/mol

mol HI = 0.011 mol

capacity of the buffer will not be exceed

3 0
3 years ago
Choose a generic equation showing how a weak base ionizes water.
FinnZ [79.3K]
Hi!

The generic equation showing how a weak base ionizes water is the following:

B(aq) + H₂O(l) ⇄ BH⁺(aq) + OH⁻(aq)

A base is described as weak when its dissociation is difficult and isn't complete. Weak Bases are chemical compounds that accept protons from water, to form the conjugate acid and the OH⁻ ion. The pH of the solution of a weak base is higher than 7. Some weak bases include ammonia (NH₃), Aluminum Hydroxide (Al(OH)₃) and Sodium Acetate (CH₃COONa).

Have a nice day!
4 0
3 years ago
What is the word equation for sodium polyacrylate and water?
DIA [1.3K]

Chemical Formula: Water Formula- H 2 O Sodium Polyacrylate Formula- (C 3 H 3 NaO 2 ) n Research Question: What is the effect of sodium polyacrylate on the temperature of the chemical mixed with water? Background Research: When sodium polyacrylate is added to water, it absorbs the moisture and turns into a solid gel.

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3 years ago
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