At STP, how many moles are in 0.880L of He gas?

Anna lives in a city that experiences high precipitation, with an average annual rainfall of 524 millimeters. It is warm all yea

gavmur [86]

Anna lives in a city that is part of the tropical climate types. It has a constantly warm weather, and thus higher humidity, and according to the annual rainfall, it is most probably a rainfall that appears seasonally, not throughout the whole year.

Tim, on the other hand, lives in a city that is part of the dry climate types. It is most probably a place that is deep into the mainland, like the cold deserts of Central Asia, where the temperatures in the summer are high, and in winter are very low. Because of the distance from the sea, the rainfall doesn't reach this places, so they are very dry, and only have symbolic amount of annual rainfall.

6 0

2 years ago

What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH

vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

$2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}$

Mass of methanol = $3.24\ \text{g}$

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass $(32.04\ \text{g/mol})$

$\dfrac{3.24}{32.04}=0.10112\ \text{moles}$

Enthalpy change for the number of moles is given by

$\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}$

$\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}$

The change in enthalpy is 12.78 kJ.

5 0

2 years ago

The shape of an atomic orbital is associated with

Diano4ka-milaya [45]

The properties of the atomic orbital are actually dependent on the quantum numbers.

size of atomic orbital: governed by the principal quantum number (n)

shape of atomic orbital: governed by the angular momentum quantum number (l)

orientation in space: governed by the magnetic quantum number (ml)

Since we are asked about the shape, hence the correct answer is:

angular momentum quantum number (l)

8 0

2 years ago

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

baherus [9]

Explanation:

Formula for work done is as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now, we will put the given values into the above formula and calculate work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is $3.68 \times 10^{-18} J$ .

7 0

2 years ago

The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3

jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 $in^{3}$

Given density = 0.242 g/ $cm^{3}$

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is $in^{3}$ and unit of volume in density is $cm^{3}$ , so first we need to change the unit of volume from $in^{3}$ to $cm^{3}$ so that the volume units will cancel out, leaving only the mass units.