Determina el número de moles y moléculas de 25 gr de HCI

Which type of stoichiometric calculation does not require the use of the molar mass?

Marianna [84]

The choices can be found elsewhere and as follows:

a. mass-mass problems
b. mass-volume problems 
c. mass-particle problems 
d. volume-volume problems

I believe the correct answer is option D. It is volume-volume problems that does not require the use of molar mass. Here you are dealing with molarities and volumes to determine concentrations. Molar mass is not part of any calculations.

6 0

3 years ago

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Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0

3 years ago

Give the quantum number set for one electron in the 3p sub level of a sulfur (S)

vesna_86 [32]

Answer:- Atomic number for sulfur is 16 and it's electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^4$ . Here, there are total for electrons in 3p and the set of quantum numbers for these 4 electrons would be as..

For the first electron of 3p-

n = 3, l = 1, ml = -1 and ms = +(1/2)

for the second electron of 3p-

n = 3, l = 1, ml = 0 and ms = +(1/2)

for the third electron of 3p-

n = 3, l = 1, ml = +1 and ms = +(1/2)

and for the fourth electron of 3p-

n = 3, l = 1, ml = -1 and ms = -(1/2)

3 0

3 years ago

Considering the temperature vs. time graph below, how does the temperature at the beginning of a change of state

Cloud [144]

Answer:

The temperature is always lower.

Explanation:

The temperature is always lower at the end of the state as compared to beginning of the state. We can see in the given data, the temperature is higher at the beginning i. e. 140 degree Celsius but with the passage of time, the temperature of a state decreases constantly and the temperature at the end is lower i. e. 20 degree Celsius. So we can conclude that the temperature is always lower.

6 0

3 years ago

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The combustion of palmitic acid is represented by the chemical equation: C16H32O2(s) + 23O2(g) → 16 CO2(g) + 16 H2O(l) The magni

solniwko [45]

Answer:

The correct option is C

Explanation:

From the question we are told that

The reaction is

$C_{16}H_{32}O_2(g) + 23O_2(g) \to 16 CO_2(g) + 16 H_2O(l)$

Generally $\Delta H = \Delta U + \Delta N* RT$

Here $\Delta H$ is the change in enthalpy

$\Delta U$ is the change in the internal energy

$\Delta N$ is the difference between that number of moles of product and the number of moles of reactant

Looking at the reaction we can discover that the elements that was consumed and the element that was formed is $O_2$ and $CO_2$ and this are both gases so the change would occur in the number of moles

So

$\Delta H = \Delta U + [16 -23]* RT$

$\Delta H = \Delta U -7RT$

The negative sign in the equation tell us that the enthalpy $\Delta_r H$ would be less than the Internal energy $\Delta_r U$

4 0

3 years ago