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olga_2 [115]
3 years ago
10

Determine the Concentration of the Unknown Strong Acid

Chemistry
1 answer:
Natasha2012 [34]3 years ago
3 0
Hey baby I know it’s late but I wanted to say how are u doing
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20 ml each of oxygen and methane are allowed to react when the reaction is completed.  The system now contains
Mekhanik [1.2K]

Answer:

CH4(g)+2O2(g)→CO2(g)+2H2O(l).

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3 years ago
A gas held at constant volume is heated from -5 degrees C to 50 degrees C, if the initial pressure is 1.0 atm, what is the new p
Hunter-Best [27]
1.21 atm, assuming the gas behaves ideally
8 0
3 years ago
A student mixes chemicals A and B together and records the amount of time it takes for a color change to occur. The student repe
k0ka [10]

Answer:

It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.

8 0
3 years ago
An ion with 11 protons 12 neutrons and a charge of + 1 has an atomic number of
posledela

Answer:

11 electrons total and 23amu

Explanation:

This tells us that sodium has 11 protons and because it is neutral it has 11 electrons. The mass number of an element tells us the number of protons AND neutrons in an atom (the two particles that have a measurable mass). Sodium has a mass number of 23amu.

5 0
3 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
Read 2 more answers
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