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Svetradugi [14.3K]
3 years ago
14

For example: The solid is not completely submerged in the water. Will this technique error increase, decrease, or have no effect

on the reported density of the solid?
Chemistry
1 answer:
Darya [45]3 years ago
8 0
Density is the amount of matter in an object so having it any way submerged in water will not affect the solid's density
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Which of the following is the best name for a compound made from calcium and bromine? (CaBr2) calcium bromide calcium dibromide
Mice21 [21]

Answer:

Calcium bromide

Explanation:

When naming compounds, the use of prefixes depend on the type of bond made. In this case, calcium and bromine form a ionic bond because calcium is a metal and bromine is a non-metal.

Ionic bonds are not named using prefixes. So no matter how many atoms there are, you will simply write the name of the element for the first element.

For the second element, you name it as well, but only use the root name and end it with -ide.

7 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
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olganol [36]

Answer: they will slow down

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3 years ago
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A company is interested in buying a new machine to replace their outdated equipment. However, before committing to the purchase,
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Efficiency = work output / work input Efficiency = (1540 J / 1600 J) x 100= 96.25%
Thus, the answer is B.
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3 years ago
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