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2 years ago

A. Stamens (p) The portion of the pistil receiving the pollen

Chemistry

2 answers:

Alja [10]2 years ago

6 0

Answer:

(A) A-(S) ; B-(R) ; C-(Q) ; D-(P).

Explanation:

(A)-S_ The male part of a flower.

(B)-R _The stalk of an anther.

(C)-Q_ The female structure of a flower.

(D)-P _The portion of the pistil recieving the pollen.

Shalnov [3]2 years ago

5 0

I believe the correct answer is B

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2 years ago

The compound methylamine, CH3NH2, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibrium

Veronika [31]

Answer:

Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]

Explanation:

According to Brönsted-Lowry acid-base theory:

An acid is a substance that donates H⁺.
A base is a substance that accepts H⁺.

When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.

CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)

The basic equilibrium constant (Kb) is:

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2 years ago

How many moles of sand (SiO2) are in 30 g of sand?

anastassius [24]

<h3>Answer:</h3>

0.50 mol SiO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

30 g SiO₂ (sand)

<u>Step 2: Identify Conversions</u>

Molar Mass of Si - 28.09 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of SiO₂ - 28.09 + 2(16.00) = 60.09 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 30 \ g \ SiO_2(\frac{1 \ mol \ SiO_2}{60.09 \ g \ SiO_2})$
Multiply/Divide: $\displaystyle 0.499251 \ mol \ SiO_2$

<u>Step 4: Check</u>

<em>Follow sig figs and round. We are given 2 sig figs.</em>

0.499251 mol SiO₂ ≈ 0.50 mol SiO₂

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2 years ago

Read 2 more answers

Why is melting an ice cube a reversible physical change?

Dmitriy789 [7]

Answer:

Because you can freeze the water you melted back into a ice cube.

Explanation:

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2 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

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2 years ago