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Free_Kalibri [48]
3 years ago
12

Which organelle is like the brain of the cell?

Chemistry
2 answers:
Lisa [10]3 years ago
6 0

Answer:

nucleus

Explanation:

UkoKoshka [18]3 years ago
3 0

Answer:

The nucleus

Explanation:

The nucleus can be thought of as the brains of a cell.

-CancerQuest

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20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as
shepuryov [24]

The reaction forms 98.76 g AlCl_3.  

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

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

<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.  

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al

Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of AlCl_3 we can obtain from each reactant.  

<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

4 0
2 years ago
A patient needs to be given exactly 1000 ml of a 5.0% (w/v) intravenous glucose solution. the stock solution is 55% (w/v). how m
LiRa [457]
The patient needs 1000 ml of 5% (w/v) glucose solution
i.e  1000 ml x 5 g/ 100 ml 
where the stock solution is 55% (w/v) = 55 g / 100 ml  
So, 1000 ml x 5 g / 100 ml = V (ml) x 55 g / 100 ml 
V = 1000 x (5 / 100) / (55 / 100) = 5000 / 55 = 90.9 ml 
∴ the patient needs 90.9 ml of 55% (w/v) glucose solution

6 0
3 years ago
The density of no2 in a 4.50 l tank at 760.0 torr and 25.0 °c is ________ g/l.
kap26 [50]
From  ideal  gas  equation   that   is   PV=nRT
n(number of  moles)=PV/RT
P=760 torr
V=4.50L
R(gas  constant =62.363667torr/l/mol
T=273 +273=298k
n  is   therefore   (760torr x4.50L) /62.36367 torr/L/mol  x298k  =0.184moles
the  molar  mass  of  NO2 is  46  therefore  density=  0.184  x  46=8.464g/l
7 0
2 years ago
Read 2 more answers
Which unit can be used to express the rate of a reaction?
vichka [17]
<span>The rate of reaction may be expressed as a unit of quantity divided by a unit of time. The only expression that has a quantity divided by time is the first one mL / s (i.e. milliliter per second), so the answer is the first option, mL/s.</span><span />
4 0
3 years ago
Read 2 more answers
Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent
Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0
3 years ago
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