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Tomtit [17]
3 years ago
9

Calculate the amount of heat needed to raise the temperature of 4 ml of water from 10C to 30C

Chemistry
1 answer:
Over [174]3 years ago
4 0

Answer:

Q=80cal

Explanation:

Hello,

In this case, the relationship between heat, mass, specific heat and change in temperature is understood by:

Q=mCp(T_2-T_1)

In this case, since the involved substance is water, whose specific heat is 1 cal/(g°C), we compute the heat 4 mL of water need to rise the temperature from 10 °C to 30 °C as shown below:

Q=4mL*\frac{1g}{1mL}*1\frac{cal}{g\°C}*(30\°C-10\°C)\\  \\Q=80cal

Best regards.

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For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?
damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.</em>

<em />

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

<em>The moles of CH3COO- are its initial moles - the moles of HCl added</em>

<em>And moles of CH3COOH are its initial moles + moles HCl added</em>

<em />

Moles CH3COO-:

Initial moles  = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles  = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

<em>pH = 3.90</em>

<em />

<h3>The pH of the buffer is 3.90</h3>
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3 years ago
A ____________ is a large body that moves around a star?
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Answer:

A Planet

Explanation:

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zalisa [80]

Answer:

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Explanation:

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During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L
Studentka2010 [4]

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:

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Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

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And we <u>solve for V</u>:

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CaHeK987 [17]

Answer:

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