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2 years ago

. A standard dry cell has an output voltage of A. 1.5 VDC. B. 1.1 VDC. C. 1.2 VDC. D. 2.0 VDC.

1 answer:

4 0

<span>The correct answer is letter A. 1.5 VDC. A standard dry cell has an output voltage of A. 1.5 VDC. Standard dry cell is a type of electricity-producing chemical cell, that is commonly use in households, and even portable devices. Dry cell is zinc-carbon cell and with nominal voltage of 1.5 volts.</span>

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Determine the molality of a solution in which 2.5 grams of glucose C12H22011 are dissolved in 1492grams of water.

Roman55 [17]

The answer is A hope that helps

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2 years ago

Which answer best describes what is happening in the following reaction?

NISA [10]

A. This is not a redox reaction. It is an example of combustion.

<h3>Combustion reaction of hydrocarbon</h3>

During the combustion of a hydrocarbon, the hydrocarbon reacts with oxygen to create carbon dioxide, water, and heat.

<h3>Example of combustion reaction</h3>

2C8H18 + 25O2 → 16CO2 + 18H2O

Thus, we can conclude that, this is not a redox reaction. It is an example of combustion.

Learn more about combustion here: brainly.com/question/9425444

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1 year 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

2 years ago

olya-2409 [2.1K]

B.) Because they both are Acids

Hope this helps!

4 0

2 years ago

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Which elements are present in the compound aluminium nitrate?

PSYCHO15rus [73]

Answer:

Aluminum nitrate is a salt composed of aluminum and nitric acid, belonging to a group of reactive chemicals - organic nitrate and nitrite compounds. The nitrate ion is polyatomic, meaning it is composed of two or more ions that are covalently bonded. This ion makes up the conjugate base of nitric acid.

Explanation:

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2 years ago

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