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liraira [26]
3 years ago
14

One of the reactions you observed resulted in this product: nacl + h2o + co2 (g)? what well did this reaction occur in? describe

how the observations for this reaction support your answer.
Chemistry
1 answer:
Nostrana [21]3 years ago
7 0
Chemical reaction: NaHCO₃ + HCl → NaCl + CO₂ + HCl.
Reaction occurs between sodium bicarbonate (sodium hydrogen carbonate) and hydrochloric acid. For example, hydrogen carbonate<span> mixed with water is used like </span>antacid (neutralizes <span>stomach acidity), contverts </span>stomach acid (hydrochloric acid) to carbon dioxide.
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When the cell splits, they have a nucleus with complete genetic material for the new daughter cells. What is this genetic materi
vaieri [72.5K]
Chromosomes is the correct answer darlin
8 0
3 years ago
Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)     \Delta H_1=-241.8kJ

(2) X(s)+2Cl_2(g)\rightarrow XCl_4(s)    \Delta H_2=+461.9kJ

(3) \frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)    \Delta H_3=-92.3kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) H_2O(g)\rightarrow H_2O(l)    \Delta H_5=-44.0kJ

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) 2H_2O(g)\rightarrow 2H_2(g)+O_2(g)     \Delta H_1=2\times 241.8kJ=483.6kJ

(2) XCl_4(s)\rightarrow X(s)+2Cl_2(g)    \Delta H_2=-461.9kJ

(3) 2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)    \Delta H_3=4\times -92.3kJ=-369.2kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) 2H_2O(l)\rightarrow 2H_2O(g)    \Delta H_5=2\times 44.0kJ=88.0kJ

The expression for enthalpy of main reaction will be:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5

\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)

\Delta H=-1048.6kJ

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0
3 years ago
Calculate the pH of the following simple solutions:
IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

<u>(1) 53.1 mM HCl</u>

Concentration of HCl = 53.1mM=53.1\times 10^{-3}M

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= 53.1\times 10^{-3}M

pH : It is defined as the negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

pH=-\log (53.1\times 10^{-3})

pH=1.27

<u>(2) 0.223 M KOH</u>

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (0.223)

pOH=0.65

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35

<u>(3) 53.1 mM HCl + 0.223 M KOH</u>

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0
3 years ago
Is oatmeal a compound​
Alika [10]

Answer:

no

Explanation:

its made of a living organisim which in this case is oats

6 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST FOR THE BEST ANSWER~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
adelina 88 [10]
There are 1,000 milligrams (mg) in one gram:
In 10 grams, there are 10 x 1,000 = 10,000 milligrams. This is a lethal dose of caffeine.

There are 4.05 mg/oz (milligrams/ounce) of caffeine in the soda. 
In a 12 ounce can, there are 4.05 x 12 = 48.6 milligrams.

How many sodas would it take to kill you?
To find this, we divide the lethal dose amount (10,000 mg) by the amount of caffeine per can (48.6 mg).
10,000 ÷ 48.6 = 205.76. 

Since 205 cans is not quite 10,000 mg, technically it would take 206 cans of soda to consume a lethal dose of caffeine. 
7 0
3 years ago
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