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Ahat [919]
3 years ago
12

When measuring the volume of the marble, why is it better to use the 25-mL graduated cylinder than the 100-mL graduated cylinder

?
Chemistry
1 answer:
AnnZ [28]3 years ago
5 0

Answer:

\boxed {\tt A \ marble \ is \ small}

Explanation:

A marble is not a very large object, so a smaller graduated cylinder is a better choice.  A 100 milliliter graduated cylinder is not needed to measure the volume of a small marble, so a 25 milliliter graduated cylinder is the best option.

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What is the concentration of a solution with a volume of 1 L that contain 200 grams of Fe(OH)3?
Goryan [66]

Answer:

0.187M

Explanation:

5 0
2 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
What is the term for the number of protons in the nucleus of each atom of an element?
xxTIMURxx [149]

Answer:

atomic number

Explanation:

atomic number is the number of protons

5 0
1 year ago
What can be said about an exothermic reaction with a negative entropy change?.
pashok25 [27]
Spontaneous at low temperatures.
3 0
2 years ago
2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.
AnnZ [28]

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+   +  C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2  + H2O ⇄  HC2O4-  +  OH-

In water the salt is basic.  The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4-  +  H2O ⇄  H2C2O4  +  OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4  + H2O ⇄ H3O+  + HC2O4-

HC2O4-  +  H2O ⇄  H3O+  C2O4-2

The  HC2O4-  acts as an ampholyte since it accepts and delivers protons simultaneously.

6 0
3 years ago
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