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KIM [24]
2 years ago
10

Net Ionic Equations for mixing Strong Acids with Strong Bases Consider a reaction between hydrochloric acid and potassium hydrox

ide.
a) In an aqueous solution of hydrochloric acid, check all of the major species found in solution (ignore the trace hydronium ions and hydroxide ions that would come from the autoionization of water). HCI (1) CIo (aq) CI (aq) OH (ag)H20 () H3o (aq)
b) In an aqueous solution of potassium hydroxide, check all of the major species found in solution (ignore the trace hydronium ons and hydroxide ions that would come from the autoionization of water). H20 () K+ (aq) OH (aq) Hyo (aq) KOH (5)
c) When the acid and base react together, they will neutralize each other to form water and a salt. Give the chemical formula for the salt formed. chemPad Help Greek acid and potassium hydroxide. Write it out in this order. Remember that, by convention, a net ionic equation has a single reaction arrow
d) When you cancel out the spectator ions, what is the net ionic equation that remains for the reaction between hydrochloric Help chemPad Oreek ▼
Chemistry
1 answer:
AleksandrR [38]2 years ago
4 0

Answer:

sorry but I am just answering the questions because I need points

Explanation:

thank you

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A doctor's Order is 0.125 g of ampicillin. The liquid suspension on hand contains 250 mg/s. How many milliliters (mL) of the sus
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Answer:

c) 2.5 mL

Explanation:

Solution

Doctors order = 0.125g

and

The liquid suspension concentration = 250 mg/5ml

= 0.250g/5ml

Or 0.05g/ml

Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml

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2 years ago
A factor that does no change in an experiment is the.
kvasek [131]

Answer:

A. Controlled variable

Explanation:

a controlled variable or a constant variable is a variable that doesnt change during an experiment

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3 years ago
5. What is the density of water in g/mL? Why?​
goldfiish [28.3K]

Answer:

1g/ml @ 4 degrees C by definition

Explanation:

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3 A partial desert food web is shown.
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Decomposeza is correct
8 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
2 years ago
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