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astra-53 [7]
3 years ago
9

How does a limiting reagent affect how much product is formed?

Chemistry
2 answers:
Anna35 [415]3 years ago
4 0

Answer: C

Explanation:

Anastasy [175]3 years ago
3 0

Answer:

<em>C. The amount of product will be limited by the number of moles of</em>

<em>the limiting reagent.</em>

Explanation:

<em>a p e x :)) i just took the quiz</em>

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Which of the following lists describes characteristics of a base? (3 points)
mel-nik [20]

Answer:

slippery, high pH, and caustic (last option)

Explanation:

when we say base we should think soap. soap is slippery. Bases give OH- ion. when OH- is combined with H+ ion it will create water which raises the pH. Since base can dissolve fats, ex: using dish liquid to cut grease on pots and pans etc.. they are caustic. Biologically they can disrupt the cell memebrane making it caustic to cell tissue.

3 0
3 years ago
Define the standard heat of formation of mgco3
Karo-lina-s [1.5K]
The standard heat of formation is the energy change that occurs when 1 mole of a substance is made from its elements in their standard states ( most stable form of the element at 1 bar of pressure and the specified temperature, usually 298 K or 25 degrees Celcius) 
E.g for MgCO3
Mg (s) + C (s) + 3/2 O2 (g) = MgCO3
7 0
3 years ago
When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
Naddik [55]

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

6 0
3 years ago
) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react
Tom [10]

<u>Answer:</u> The amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of B_5H_9 with oxygen gas follows:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ

To calculate the amount of heat released for the given amount of B_5H_9(l), we use unitary method, we get:

When 2 moles of B_5H_9(l) reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of B_5H_9(l) will react, the amount of heat released will be = \frac{5259}{2}\times 0.211=554.8kJ

Hence, the amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

7 0
3 years ago
2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.
Savatey [412]
<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

Glutamic -----> 3.08

Asparagine -----> 5.43

Tyrosine -----> 5.63

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

5 0
2 years ago
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