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Firdavs [7]
3 years ago
11

What's the pollen tube also known as

Chemistry
1 answer:
kkurt [141]3 years ago
8 0

Answer:

The stigma.

Explanation:

You're welcome. :)

You might be interested in
Please help I really appreciate it thank you ❤️
Feliz [49]

II. sulfur (S) and carbon (C)

and

III. fluorine (F) and oxygen (O)

will form covalent bonds, so the answer will be:

e. II and III

Explanation:

To know is what type of bond is formed between atoms we need to look at the electronegativity difference between the atoms.

If the electronegativity difference is less than 0.4 there is a nonpolar covalent bond.

If the electronegativity difference is between 0.4 and 1.8 there is a polar covalent bond. (if is a metal involved we consider the bond to be ionic)

If the electronegativity difference is greater then 1.8 there is an ionic bond.

We have the following cases:

I. lithium (Li) and sulfur (S)

electronegativity difference = 2.5 (S) - 1 (Li) = 1.5 but because there is a metal involved the bond will be ionic

II. sulfur (S) and carbon (C)

electronegativity difference = 2.5 (S) - 2.5 (C) = 0 so the bond will be nonpolar covalent

III. fluorine (F) and oxygen (O)

electronegativity difference = 4 (F) - 3.5 (O) = 0.5 so the bond will be polar covalent bond.

Learn more about:

covalent and ionic bonds

brainly.com/question/1802971

#learnwithBrainly

3 0
4 years ago
A 5.0 L sample of gas at 300. K is heated to 600. K. What will the new volume of the gas be?
Ainat [17]

Answer:

V_2=10L

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required new volume by using the Charles' law as a directly proportional relationship between temperature and volume:

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

In such a way, we solve for V2 and plug in V1, T1 and T2 to obtain:

V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{5.0L*600K}{300K}\\\\V_2=10L

Regards!

4 0
3 years ago
Strong acids have a ______ value of Ka and the equlibrium for the reaction with water lies far to the _______.
nika2105 [10]

Answer:

All strong acids have a higher value of K_{a} and the equilibrium for the reaction with water lies far to the right.

Explanation:

All strong acids dissociate completely in the solution. Higher the value of dissociation constant of the acid, higher will be the dissociation of the acid.

The reaction of the acid with water will be favored in the forward direction for acids having higher dissociation constant value (K_{a}).

The dissociation of a strong acid say HA in water is shown below\textrm{HA}\left ( aq \right )+\textrm{H}_{2}\textrm{O}\left ( l \right )\rightleftharpoons \textrm{A}^{-}\left ( aq \right )+\textrm{H}_{3}\textrm{O}^{+}\left ( aq \right )

Higher the value of K_{a}, more will be the dissociation of the acid in water. The reaction will move far to the right side.

4 0
3 years ago
Is air pollution a chemical change?
Evgesh-ka [11]

Answer: many of the chemicals found in urban air are formed by chemical reactions driven by sunlight.

4 0
3 years ago
Read 2 more answers
A gas occupies 3.8 L at -18° C and 975. torr. What volume would this gas occupy at STP?
Aleks04 [339]

To solve the question we will assume that the gas behaves like an ideal gas, that is to say, that there is no interaction between the molecules. Assuming ideal gas we can apply the following equation:

PV=nRT

Where,

P is the pressure of the gas

V is the volume of the gas

n is the number of moles

R is a constant

T is the temperature

Now, we have two states, an initial state, and a final state. The conditions for each state will be.

Initial state (1)

P1=975Torr=1.28atm

V1=3.8L

T1=-18°C=255.15K

Final state(2), STP conditions

P2=1atm

T2=273.15K

V2=?

We will assume that the number of moles remains constant, so the nR term of the first equation will be constant. For each state, we will have:

\begin{gathered} \frac{P_1V_1}{T_1}=nR \\ \frac{P_2V_2}{T_2}=nR \end{gathered}

Since nR is the same for both states, we can equate the equations and solve for V2:

\begin{gathered} \frac{P_{2}V_{2}}{T_{2}}=\frac{P_1V_1}{T_1} \\ V_2=\frac{P_{1}V_{1}}{T_{1}}\times\frac{T_2}{P_2} \end{gathered}

We replace the known values:

V_2=\frac{1.28atm\times3.8L}{255.15K}\times\frac{273.15K}{1atm}=5.2L

At STP conditions the gas would occupy 5.2L. First option

8 0
1 year ago
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