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3 years ago

What is the independent and dependent variable

Chemistry

1 answer:

MakcuM [25]3 years ago

8 0

Independent- controlled in a scientific experiment to test the effects a dependent is being tested and measured in a scientific experimrent

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How many moles of H2O form when 4.5 moles O2 reacts?

mart [117]

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.

5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2

Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.

4 0

3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

Question 11. Identify the reducing agent <br> Sn+2 + AG 0 —> Sn0 + Ag+

Temka [501]

Answer:

Ag 0 is the reducing agent.

Explanation:

Reducing -> gaining electrons

Oxidizing -> losing electrons

Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.

4 0

2 years ago

What are the functions of ammonia in the solvay process

enot [183]

The ammonia gas is absorbed in the concentrated brine to produce aqueous sodium chloride and aqueous ammonia. This ammoniation process is exothermic, so energy is released as heat. The ammonia tower eventually needs to be cooled.

7 0

3 years ago

Anyone know this please help me

natka813 [3]

Answer:go to the sight called would or do help

Explanation:go onnnnnnn

6 0

3 years ago