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Burka [1]
3 years ago
13

For example," "however," and "also" are examples of:

Chemistry
2 answers:
Vera_Pavlovna [14]3 years ago
7 0
Conjunctions, A. You’re welcome
sladkih [1.3K]3 years ago
6 0

Answer:

conjunctive adverb...................

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PLEASE PLEASE PLEASE HELP! WILL AWARD BRAINLIESTPLEASE ANSWER AND STOP JUST LOOKING AT ITTTTT
Goshia [24]

Answer:

answer - c

answer - a

Explanation:

<h2>I hope answer correct</h2><h2><em><u>pl</u></em><em><u>ease</u></em><em><u> like</u></em><em><u> me</u></em></h2>
3 0
3 years ago
How many moles of ZnCl, will be produced from 23.0 g of Zn, assuming CuCl, is available in excess?
Dennis_Churaev [7]

Answer:

23 g of Zn will produce 0.352 mole of zinc chloride.      

Explanation:

Given data:

mass of Zn = 23 g

Moles of ZnCl = ?

Solution:

Chemical equation:

2Zn + CuCl₂ → 2ZnCl + Cu

Number of moles of Zn:

<em>Number of moles of Zn = mass/ molar mass</em>

Number of moles of Zn = 23 g / 65.38 g/mol

Number of moles of Zn = 0.352 mol

now we will compare the moles of ZnCl with Zn because it is limiting reactant.

                         Zn       :      ZnCl  

                         2         :        2

                         0.352 :        0.352

So 23 g of Zn will produce 0.352 mole of zinc chloride.      

7 0
4 years ago
I need someone to help me with this. PLEASE, I'M BEGGING YOU
musickatia [10]

Answer:

D. 25%

Explanation:

8 0
3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

4 0
3 years ago
A negative ΔH means an exothermic reaction. <br> True <br> False
slega [8]
True a negative change in enthalpy means an exothermic reaction
7 0
3 years ago
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