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2 years ago

Tyrone wants to buy peanut butter that does not contain any hydrogenated

Chemistry

1 answer:

ryzh [129]2 years ago

8 0

The basic issue that Tyrone be concerned about is that there are unknown risks to not using hydrogenated products.

<h3>What is a hydrogenated oil?</h3>

A hydrogenated oil is one which is produced from fat by passing hydrogen gas through the fat at high temperature over a catalyst.

The basic issue that Tyrone be concerned about is that there are unknown risks to not using hydrogenated products.

Learn more about fats and oils:brainly.com/question/11587174?

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Why is it appropriate that galileo galilei is known as a leader in the study of space

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He was actually the first one to look at the moon.

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3 years ago

What are some ideas for a compound machine for my middle school invention science project that can be useful in people's daily l

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A toothbrush that has toothpaste in the middle so it comes preloaded. That way it is cleaner and uses more product than normal

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4 years ago

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.57 . A. Determine

vlada-n [284]

Answer:

[C₆H₅NH₃⁺] = 0.0399 M

Explanation:

This excersise can be easily solved by the Henderson Hasselbach equation

C₆H₅NH₃Cl → C₆H₅NH₃⁺ + Cl⁻

pOH = pKb + log (salt/base)

As we have value of pH, we need to determine the pOH

14 - pH = pOH

pOH = 8.43 (14 - 5.57)

Now we replace data:

pOH = pKb + log ( C₆H₅NH₃⁺/ C₆H₅NH₂ )

8.43 = 9.13 + log ( C₆H₅NH₃⁺ / 0.2 )

-0.7 = log ( C₆H₅NH₃⁺ / 0.2 )

10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2

0.19952 = C₆H₅NH₃⁺ / 0.2

C₆H₅NH₃⁺ = 0.19952 . 0.2 = 0.0399 M

5 0

3 years ago

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

Given:

Molar mass of $H_2SO_4$ = 98 g/mole

Mass of $H_2SO_4$ = $6.05\times 10^3kg=6.05\times 10^6g$

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

$\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg$

Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

3 0

3 years ago

for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3

Y_Kistochka [10]

Answer:

2 mole MnO₂

Explanation:

2Mn(s) + 2O₂(g) => 2MnO₂(s)

4 0

4 years ago

Read 2 more answers