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Ksju [112]
3 years ago
14

The reaction nacl(s) → nacl(aq) is performed in a coffee cup calorimeter, using 100 ml of h2o(l) and 5.00g of nacl. if the tempe

rature of solution after mixing decreased by 2.30°c, what is the δhrxn in kj/mol? assume the heat capacities of all solutions are 4.18 j/g°c, and densities of all solutions are 1.00 g/ml. the molar mass of nacl is 58.44 g/mol
Chemistry
1 answer:
MrRissso [65]3 years ago
5 0

Since the density of water is 1 g /mL, hence there is 100 g of H2O. So total mass is:

m = 100 g + 5 g = 105 g

 

=> The heat of reaction can be calculated using the formula:

δhrxn = m C ΔT

where m is mass, C is heap capacity and ΔT is change in temperature = negative since there is a decrease

 

δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)

δhrxn = -1,009.47 J

 

=> However this is still in units of J, so calculate the number of moles of NaCl.

 

moles NaCl = 5 g / (58.44 g / mol)

moles NaCl = 0.0856 mol

 

=> So the heat of reaction per mole is:

δhrxn = -1,009.47 J / 0.0856 mol

δhrxn = -11,798.69 J/mol = -11.8 kJ/mol

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Answer:

Given

mass of H2O (m) =35.6g

molarmass (mr) = H2O ), 1x2+16=18g/mol

moles of H2O (n) =?

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n=35.6g/18g/mol

n=1.978moles

the moles of H2O are 1.978moles

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What is the conjugate base<br> of H2O?
kolbaska11 [484]

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A conjugate acid, is a species formed by the reception of a proton (H+) by a base—in other words, it is a base with a hydrogen ion added to it.

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3 years ago
If a sample of HF gas at 694.9 mmHg has a volume of 3.463 Land the volume is changed to 5.887 L, then what will be the new press
VARVARA [1.3K]

Answer:

\large \boxed{\text{381.7 mmHg}}

Explanation:

Data:

p₁ = 694.9 mmHg; V₁ = 3.463 L

p₂ = ?;                     V₂ = 5.887 L

Calculation:

\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{648.9 mmHg} \times \text{3.463 L} & = & p_{2} \times\text{5.887 L}\\\text{2247.1 mmHg} & = & 5.887p_{2}\\p_{2} & = & \dfrac{\text{2247.1 mmHg}}{5.887}\\\\& = &\textbf{381.7 mmHg}\\\end{array}\\\text{The new pressure of the gas is $\large \boxed{\textbf{381.7 mmHg}}$}

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Choose two of the following scientists: Anton Lavoisier, John Dalton, JJ Thomson, Robert Millikan, Ernest Rutherford, James Chad
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Ernest Rutherford

J. J Thomson

Explanation:

<u>Ernest Rutherford</u>

In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.

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 In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.

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To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.

<u>J. J Thomson</u>

Experiment

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Discovery and reflection on the atomic theory

From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:

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learn more:

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