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3 years ago

Which of the following is the correct meaning of transpiration?

Chemistry

2 answers:

alexdok [17]3 years ago

6 0

Answer:

The answer of this question is , The evaporation of water through the stomata.

Explanation:

Transpiration is the process by which moisture is carried through plants from roots to small pores on the underside of leaves, where it changes to vapor and is released to the atmosphere. Transpiration is essentially evaporation of water from plant leaves. Transpiration is important because water is needed for photosynthesis and because water cools a plant off.

AlladinOne [14]3 years ago

5 0

contain chlorophyll
absorb light energy
control the size of stomata
split water into hydrogen and oxygen

You might be interested in

What about 50 g of water? I need help what this

ELEN [110]

Answer:

3.38 Tablespoons

10.14 Teaspoons

0.21 U.S. Cups

0.18 Imperial Cups

0.20 Metric Cups

50.00 Milliliters

Explanation:

3 0

3 years ago

Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. N 3−, O 2

love history [14]

Answer:

Option 1: N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²

Explanation:

Ionic radius is the radius of an atom´s ion in ionic crystal structure. In an ion that lose an electron, to form a cation, the radius of the ion gets smaller, because the repulsion between electrons decrease because fewer electrons are present. Conversely, adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.

Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers.

In a isolelectronic series (same number of electrons), the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.

We will use this principle to solve our problem.

In our case, the given ions are:

N⁻³ : Z = 7, e⁻ = 10
O⁻²: Z= 8, e⁻ =10
F⁻: Z = 9, e⁻ = 10
Na⁺: Z= 11, e⁻ = 10
Mg⁺²: Z=12, e⁻ =10

where Z= number of protons, and e⁻ = number of electrons.

Hence the decreasing order of ionic radius is:

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²

Have a nice day!

4 0

2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

3 0

3 years ago

Which of the following are not single-displacement reactions?

Goshia [24]

Answer:

$\boxed{\text{B and C }}$

Explanation:

In a single-displacement reaction, one element exchanges partners with another element in a compound.

$\textbf{A. } \rm Fe + 2HCl \longrightarrow FeCl_2 + H_2$

This is a single-displacement reaction, because the element Fe exchanges partners with H in HCl.

$\textbf{B. } \rm KOH + HNO_3 \longrightarrow H_2O + KNO_3$

This is not a single-displacement reaction, because it is a reaction between two compounds.

This is a double displacement reaction in which the K⁺ and H⁺ cations change partners with the anions.

$\textbf{C. } \rm Na_2S + 2HCl \longrightarrow 2NaCl + H_2S$

This is not a single-displacement reaction. It is another double displacement reaction, in which the Na⁺ and H⁺ cations change partners with the anions.

$\textbf{D. } \rm Ca + 2HOH \longrightarrow Ca(OH)_2 + H_2$

This is a single-displacement reaction, because the element Ca exchanges partners with H in H₂O.

$\boxed{\textbf{B and C }}$ are not single-displacement reactions.

6 0

3 years ago