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3 years ago

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When pollen from a water lily combines with the egg of another water lily, it creates a seed. What is the name of this plant pro

cess?
fertilization

germination

seed dispersal

pollination

2 answers:

pickupchik [31]3 years ago

8 0

Answer:

fertilization

Explanation:

Brut [27]3 years ago

6 0

Answer:

fertilization

Explanation:

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Identify the acids and the bases in the chemical equation

agasfer [191]

<h3><u>Answer;</u></h3>

Acids; HCl and HC5H5N+

Bases; C5H5N and Cl-

<h3><u>Explanation;</u></h3>

According to Bronsted-Lowry Theory an acid is a proton or hydrogen ion donor while a base is a proton or a hydrogen ion acceptor.
In this case,<u> both HCl and HC5H5N+ are acids</u> as <u>they are donors of hydrogen ions</u>. HCl is an acid to the forward reaction while HC5H5N+ is a acid to the reverse reaction.
On the other hand, <u>C5H5N and Cl- are bases</u>, <u>they are acceptors of hydrogen ions</u>. Cl- is a base in the reverse reaction while C5H5N is a base in the forward reaction.

8 0

3 years ago

How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102

Ede4ka [16]

Answer: $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles$

1 mole of hydrogen $(H_2)$ = $2\times 6.023\times 10^{23}=12.05\times 10^{23}$ atoms

17.5 mole of hydrogen $(H_2)$ = $\frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25}$ atoms

There are $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0

3 years ago

Suppose you are a food chemist working for a company that makes and manufactures soda. Your job is to create a new soft drink wi

Mekhanik [1.2K]

Answer:

The answer to your question is given after the questions so I just explain how to get it.

Explanation:

a)

Get the molecular weight of Phosphoric acid

H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)

= 3 + 31 + 64

= 98 g

98 g ----------------- 1 mol

0.045 g --------------- x

x = (0.045 x 1) / 98

x = 0.045 / 98

x = 0.00046 moles or 4.6 x 10 ⁻⁴

b)

Molarity = $\frac{moles}{volume}$

Molarity = $\frac{0.00046}{0.35}$

Molarity = 0.0013 or 1.31 x 10⁻³

c)

Formula C₁V₁ = C₂V₂

V₁ = C₂V₂ / C₁

Substitution

V₁ = (0.0013)(1) / 0.01

Simplification and result

V₁ = 0.0013 / 0.1

V₁ = 0.13 l = 130 ml

7 0

3 years ago

torisob [31]

Decrease

Increase

Increase

Decrease

Increase

Decrease

Hope this helps! :)

4 0

3 years ago

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0

3 years ago