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madam [21]
4 years ago
11

a titration of 25.0mL of an x M hcl solution with 0.15M NaOH starts at a burette reading for NaOH of 0.20mL . Thr burette readin

g of the end point is 24.10mL. what is the volume of NaOH dispensed?
Chemistry
1 answer:
VARVARA [1.3K]4 years ago
6 0
Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL. 
Small volumes of solutions can be transferred from the burette at a controllable rate. 
In this instance NaOH is in the burette.
Initial reading of NaOH is 0.20 mL 
end point is the point at which the chemical reaction reaches completion. In acid base reactions, end point is when all the H⁺ ions have reacted with OH⁻ ions. 
final reading of NaOH is 24.10 mL
to find the volume of NaOH dispensed we have to find the difference between final reading and initial reading 
volume of NaOH added = 24.10 mL - 0.20 mL = 23.90 mL
volume of NaOH dispensed is 23.90 mL
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Answer:

Bin 1 points to a carbon bonded to a double bonded carbon and single bonded to two hydrogens. --- trigonal planar, tetrahedral

Bin 2 points to a carbon double bonded to a carbon and single bonded to a carbon and one hydrogen.------- trigonal planar, tetrahedral

Bin 3 is a carbon single bonded to two carbons and single bonded to two hydrogens. ----- tetrahedral, tetrahedral

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Explanation:

A single C-C or C-H bond is in a tetrahedral geometry, the carbon atom is bonded to four species with a bond angle of 109°.

A C=C bond is trigonal planar with a bond angle of 120°.

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3 years ago
What cell structure reminds me/you of hand sanitizer A. golgi apparatus B. endoplasmic reticulum C. cytoplasm
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3 years ago
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Answer:

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Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

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