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3 years ago

Suggest an environment where plants would not need a cuticle

Chemistry

1 answer:

Marta_Voda [28]3 years ago

7 0

Maybe the dessert, because it's dry

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What are differences between vaccines and antibotics

lara31 [8.8K]

Answer:

Explanation:

Antibiotics are chemical agents that are used to treat bacterial infections. ... Vaccines are used to prevent infection, particularly viral infections.

3 0

3 years ago

Read 2 more answers

Suppose 20.23 g of glucose are dissolved in 95.75 g of water at 27.0 OC. Glucose is nonvolatile (has no vapor pressure) and has

leonid [27]

Answer:

Explanation:

From the information given :

we can understand the solute is glucose and the solvent is water,

So, the weight of glucose = 20.23 g

the molecular weight of glucose = 180.2 g/mol

weight of water = 95. 75 g

the molecular weight of water = 18.02 g/mol

pure vapor pressure of water $P_A = 26.7 \ mmHg$ at 27°C

moles of glucose = weight of glucose/ molecular weight of glucose

= 20.23/180.2

= 0.11 mole

moles of water = weight of water / molecular weight of water

= 95.75/18.02

= 5.31 mole

mole fraction of glucose $X_{glucose} =$ (moles of glucose)/(moles of glucose+ moles of water)

$X_{glucose} =$ 0.11/(0.11 + 5.31)

$X_{glucose} =$ 0.0203

mole fraction of glucose $X_{water} =$ (moles of water)/(moles of water+ moles of glucose)

$X_{water} =$ 5.31/ (5.31 + 0.11)

$X_{water} =$ 0.9797

Using Raoult's Law:

$P_S = P^0_A \times X_A \ \ \ OR \ \ \ P_A = P^0_A \times X_A$

where:

$P_S$ = vapor pressure of the solution

$P_A$ = total vapor pressure of the solution

$P^0_A$ = vapor pressure of the solvent in the pure state

$X_A$ = mole fraction of solvent i.e. water

$P_A =$ 95.75 × 0.9797

$P_A =$ 93.81 mmHg

the total vapor pressure of the solution = 93.81 mmHg

4 0

3 years ago

Rank the following elements by electron affinity 1) fluorine 2) selenium 3) arsenic 4) potassum 5) argon

denis23 [38]

Answer:

Fluorine > Selenium > Arsenic > Potassium > Argon

Explanation:

Electron affinity describes the ability or readiness or tendency of an atom to gain an electron.

The higher the value, the higher the tendency. Electron affinity depends on the on the nuclear charge and atomic radius. When nuclear charge is more, electron affinity is high, when atomic radius increases electron affinity reduces.

Noble gases such as Helium, Neon, and Argon would have 0 affinity for electrons because of their stable electronic configuration. From the list, Ar is the least in terms of electron affinity.

Potassium is a metal with large electropositivity which describes the tendency of an atom to lose electrons. Potassium would readily lose electrons instead of gaining.

Between Arsenic and Selenium: Arsenic belongs to group V and Selenium group VI. The two elements both belong to period IV on the periodic table. Across a period, electron affinity increases due to increase in nuclear charge. Therefore, Selenium would have a greater electron affinity compared to Arsenic.

Fluorine has the highest electron affinity of all. It needs just an electron to complete its octet.

7 0

4 years ago

Elements donate 2 electron to produce a cation with a 2+ charge

hammer [34]

Answer: alkaline earth metals (group-IIA)

Explanation:

The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.

Alkaline earth metals donate 2 valence electrons to acquire noble gas configuration.

For example: Berrylium is the first alkaline earth metal with atomic number of 4 and thus has 4 electrons

Electronic configuration of berrylium:

$[Be]:4:1s^22s^2$

Berrylium atom will loose two electrons to gain noble gas configuration and form berrylium cation with +2 charge.

$[Be^{2+}]:2:1s^2$

Thus Elements donate 2 electron to produce a cation with a 2+ charge are alkaline earth metals.

5 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago