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Ilya [14]
3 years ago
13

Calculate the density of the red block. Round your answer to the nearest hundredth.

Chemistry
1 answer:
Nitella [24]3 years ago
6 0

Explanation:

I need more information to answer your question.

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A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/m
Ahat [919]

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70} = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{8.70\ g}{39.997\ g/mol}

Moles\ of\ NaOH= 0.2175\ mol

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{100\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.5508\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}

Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377

3 0
3 years ago
A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
OLEGan [10]

Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

6 0
3 years ago
Water has a density of 1.0 g/ml. which of these objects will float in water? object i: mass = 50.0 g; volume = 40.2 ml object ii
Oksana_A [137]
<span>Object I would be it</span>
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3 years ago
The ____________ ___________ is the outer part of the atom?
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Possibly electron shells, t<span>he </span>electrons<span> in the outermost occupied </span>shell<span> (or </span>shells<span>) determine the chemical properties of the atom..</span>
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3 years ago
Kendall wants to determine if there is a trend in air temperature changes during April . Which procedure should she chose?
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I think the answer is A
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