All categories

3 years ago

Which type of plant is this? Non-vascular Moss Perennial Vascular

Chemistry

2 answers:

Sergio [31]3 years ago

8 0

Perennial.? I think...

aivan3 [116]3 years ago

6 0

That looks like an Acacia tree on the savanna’s of Africa. Perennial plants live for many years. I think this tree survives for many years. Read about that type of tree to see if you come to that conclusion.

You might be interested in

In the following reaction, what is the quantity of heat (in kJ) released when 5.87 moles of CH₄ are burned?

IRISSAK [1]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

The enthalpy of a chemical reaction as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

In this case, the balanced reaction is:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g)

and the enthalpy reaction ∆H° has a value of -802 kJ/mol.

This equation indicates that when 1 mole of CH₄ reacts with 2 moles of O2, 802 kJ of heat is released.

When 5.87 moles of CH₄ are burned, then you can apply the following rule of three: if 1 mole of CH₄ releases 802 kJ of heat, 5.87 moles of CH₄ releases how much heat?

$heat=\frac{5.87 molesof CH_{4}x802 kJ}{1 mol of CH_{4} }$

Finally, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

Learn more:

5 0

2 years ago

Give two possible sets of four quantum numbers for the electron in hydrogen when it is in its lowest energy state.

konstantin123 [22]

Answer:

Explanation:

The electron in the lowest energy state will be found in 1 s energy level.

set of 4 possible quantum numbers

Principal quantum no : n = 1 ,

Azimuthal quantum no l = 0

Magnetic quantum no m = 0

Spin quantum no s = + 1/2

set of other quantum nos

Principal quantum no : n = 1 ,

Azimuthal quantum no l = 0

Magnetic quantum no m = 0

Spin quantum no s = - 1/2

4 0

3 years ago

Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu

Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

$r = \dfrac{a}{2\sqrt{2}}$

where;

a = edge length.

Making "a" the subject of the formula:

$a = 2 \sqrt{2} \times r$

$a = 2 \times 1.414 \times 135 \ pm$

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

$density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}$

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

$mass\ of\ atom =\dfrac{ atomic \ mass}{Avogadro \ number}$

$mass\ of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}$

Recall also that number of atoms in a unit cell for a face-centered cubic = 4

So;

$mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4$

mass of unit cell = 6.83380375 × 10⁻²² g

$Density \ of \ unit \ cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}$

Density of unit cell ( rhodium) = 12.279 g/cm³

3 0

3 years ago

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

What causes differences in air pressure

shtirl [24]

Unequal heating of the atmosphere

5 0

3 years ago

Read 2 more answers