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Goshia [24]
3 years ago
10

The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st

ems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0269 m solution of oxalic acid?
Chemistry
1 answer:
VLD [36.1K]3 years ago
3 0
Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.

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Please help me guys to solve this problem!​
steposvetlana [31]

(a) distance measured with metre rule or tape rule

    time measured with stopwatch/ stop clock/ timer

    Average speed = total distance / time

(b) (i) decrease in speed

    (ii) Change in speed = a * t

        4.5 m/s

<em>hope this helps......</em>

6 0
3 years ago
(a) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm3. (b) A student needs 15.0 g of ethanol fo
o-na [289]

<u>Answer:</u>

<u>For a:</u> The density of the sample of copper is 8.96g/cm^3

<u>For b:</u> The volume of ethanol needed is 19.0 mL

<u>For c:</u> The mass of mercury is 340. grams

<u>Explanation:</u>

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}      ......(1)

  • <u>For a:</u>

Mass of copper = 374.5 g

Volume of copper = 41.8cm^3

Putting values in equation 1, we get:

\text{Density of copper}=\frac{374.5g}{41.8cm^3}\\\\\text{Density of copper}=8.96g/cm^3

Hence, the density of the sample of copper is 8.96g/cm^3

  • <u>For b:</u>

Mass of ethanol = 15.0 g

Density of ethanol = 0.789 g/mL

Putting values in equation 1, we get:

0.789g/mL=\frac{15.0g}{\text{Volume of ethanol}}\\\\\text{Volume of ethanol}=\frac{15.0g}{0.789g/mL}=19.0mL

Hence, the volume of ethanol needed is 19.0 mL

  • <u>For c:</u>

Volume of mercury = 25.0 mL

Density of mercury = 13.6 g/mL

Putting values in equation 1, we get:

13.6g/mL=\frac{\text{Mass of mercury}}{25.0mL}\\\\\text{Mass of mercury}=(13.6g/mL\times 25.0mL)=340.g

Hence, the mass of mercury is 340. grams

5 0
3 years ago
The questions are attached​
Sladkaya [172]

Answer:

Explanation:

a) Balanced chemical equation:

C₃H₈ + 5O₂      →        3CO₂ + 4H₂O

b) Mass of oxygen gas needed = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/  44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of oxygen and propane:

                            C₃H₈        :          O₂        

                                1           :            5

                             0.57        :         5/1×0.57 =  2.85 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 2.85 mol × 32 g/mol

Mass = 91.2 g    

c) water vapors produced = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/  44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of water and propane:

                            C₃H₈        :          H₂O        

                                1           :            4

                             0.57        :         4/1×0.57 =  2.28 mol

Mass of water vapors:

Mass = moles × molar mass

Mass = 2.28 mol × 18 g/mol

Mass = 41.0 g

d)

mass of water vapors produced = 1.5 kg (1500 g)

Mass of propane reacted = ?

Solution:

Number of moles of water vapors:

Number of moles = mass/molar mass

Number of moles = 1500 g/ 18 g/mol

Number of moles = 83.33 mol

now we will compare the moles of water and propane.

         

                            H₂O          :            C₃H₈    

                               4            :              1

                             83.3         :         1/4×83.3 =  20.8 mol

Mass of propane:

Mass = number of moles ×molar mass

Mass = 20.8 mol × 44.1 g/mol

Mass = 917 g

3 0
3 years ago
Rank the following alkenes in order of increasing stability of the double bond towards addition of HBr:
ASHA 777 [7]

Answer:

2,3-dimethyl-2-butene > 3-methyl-3-hexene > cis-3-hexene > 1-hexene

Explanation:

According to Saytzeff rule, the more highly substituted an alkene is, the more stable it is. Since this is so, 2,3-dimethyl-2-butene will be the most stable of all the alkenes listed because it is the most substituted alkene.

Let us also note that terminal alkenes are the least stable because the pi bonds of the alkenes are least stabilized by alkyl groups. This implies that 1-hexene is the least stable alkene among the listed alkenes.

6 0
3 years ago
Balance the equation and identify the type of reaction for ? P4(s) + ? Ca(s) → ? Ca3P2(s). 1. 2; 6; 2 — decomposition 2. 2; 6; 2
pishuonlain [190]

Answer:

4. 1; 6; 2 — synthesis

Explanation:

<u>Decomposition reaction </u>is defined as the reaction in which a single large substance breaks down into two or more smaller substances.

AB\rightarrow A+B

<u>Single displacement </u>reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

A+BC\rightarrow AC+B

<u>Synthesis reaction</u> is defined as the reaction in which smaller substances combine in their elemental state to form a larger substance.

A+B\rightarrow AB

The unbalanced combustion reaction is shown below as:-

P_4+Ca\rightarrow Ca_3P_2

On the left hand side,  

There are 4 phosphorus atoms and 1 calcium atom

On the right hand side,  

There are 2 phosphorus atoms and 3 calcium atoms

Thus,  

Right side, Ca_3P_2 must be multiplied by 2 to balance phosphorus.

Left side, Ca is multiplied by 6 so to balance the whole reaction.

Thus, the balanced reaction is:-

P_4+6Ca\rightarrow 2Ca_3P_2

Thus, answer:- 4. 1; 6; 2 — synthesis

5 0
3 years ago
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