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2 years ago

6

Direction: To compare and differentiate the animal and plant cell base on the presence and absence of certain organelles.Fill in

the corresponding organelles in the Venn Diagram
please please answer this question properly i need it!!!!!!

1 answer:

S_A_V [24]2 years ago

3 0

Answer:

hi what are you doing

Explanation:

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victus00 [196]

Answer:

34.8 g

Explanation:

Answer:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 123.90 70.91 208.24

P₄ + 20Cl₂ ⟶ 4PCl₅

Mass/g: 46.0 32.0

2. Calculate the moles of each reactant

$\text{moles of P}_{4} = \text{46.0 g P}_{4} \times \dfrac{\text{1 mol P}_{4}}{\text{123.90 g P}_{4}} = \text{0.3713 mol P}_{4}\\\\\text{moles of Cl}_{2} = \text{32.0 g Cl }_{2} \times \dfrac{\text{1 mol Cl }_{2}}{\text{70.91 g Cl }_{2}} = \text{0.4513 mol Cl }_{2}$

3. Calculate the moles of PCl₅ we can obtain from each reactant

From P₄:

The molar ratio is 4 mol PCl₅:4 mol P₄

$\text{Moles of PCl}_{5} = \text{0.3713 mol P}_{4} \times \dfrac{\text{4 mol PCl}_{5}}{\text{4 mol P}_{4}} = \text{0.3713 mol PCl}_{5}$

From Cl₂:

The molar ratio is 4 mol PCl₅:20 mol Cl₂

$\text{Moles of PCl}_{5} = \text{0.4513 mol Cl}_{2}\times \dfrac{\text{4 mol PCl}_{5}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol PCl}_{5}$

4. Identify the limiting and excess reactants

The limiting reactant is chlorine, because it gives the smaller amount of PCl₅.

The excess reactant is phosphorus.

5. Mass of excess reactant

(a) Moles of P₄ reacted

The molar ratio is 1 mol P₄:20 mol Cl₂

$\text{Moles reacted} = \text{0.4513 mol Cl}_{2} \times \dfrac{\text{4 mol P}_{4}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol P}_{4}$

(b) Mass of P₄ reacted

$\text{Mass reacted} = \text{0.090 26 mol P}_{4} \times \dfrac{\text{123.90 g P}_{4}}{\text{1 mol P}_{4}} = \text{11.18 g P}_{4}$

(c) Mass of P₄ remaining

Mass remaining = original mass – mass reacted = (46.0 - 11.18) g = 34.8 g P₄

4 0

3 years ago

0.0400 M solution of HClO2, pH = 1.80 solve for Ka

umka2103 [35]

Ka for weak acids is ---> Ka= (X)²/ (initial- X)

initial= 0.0400 M

to find X, you can find the H concentration using pH

[H+]= 10^-pH
[H+]= 10^-1.80= 0.016 which is X

ka = (0.016)²/ (0.0400 - 0.016) = 0.011

6 0

4 years ago

Given a force of 88 N and an acceleration of 4 m/s2, what is the mass? *

pantera1 [17]

22. Mass=Force/acceleration or M=F/a. So the force of this equation would be 88N and the acceleration would be 4m/s^2 so the mass in the problem would be 22.

7 0

3 years ago

Which is the best analogy of a wave?

o-na [289]

T h e a s w e r i s d

8 0

3 years ago

3.5 liters of 0.4M HCI

nignag [31]

Answer: 1.4 moles

Explanation:

I can only assume you are looking for the amount of moles in 0.4M. the capital M means Molarity.

Molarity=moles of solute/liters of solution

Since we know the molarity is 0.4, we can plug this into our equation

$0.4M=\frac{xmoles}{3.5L}$

moles= 1.4

7 0

3 years ago