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algol [13]
3 years ago
6

1) 50.0 mL of water is measured out in a graduated cylinder. A rock with the mass of

Chemistry
1 answer:
SashulF [63]3 years ago
7 0
  • Volume=Rised height-Measured height=60-50=10mL
  • Mass=20.3g

\\ \tt\longmapsto Density=\dfrac{Mass}{Volume}

\\ \tt\longmapsto Density=\dfrac{20.3}{10}

\\ \tt\longmapsto Density=203g/mL

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if 3.26 g is dissolved in enough water to make exactly 323 ml of solution, what is the molar cocentration of nitrate ion g
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Divide both sides to get 104 and that’s your answer
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How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106​
il63 [147K]

Hey there :)

<em>Q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em> </em><em>How many km are in 5.6mm? </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>3 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-6 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-3 </em>

<em>=</em><em>></em><em> </em><em>5.6x10</em><em>^</em><em>6</em>

<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em>-</em>

5.6 \times 10^{ - 6}

<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em>-</em>

By using the formula-

1millimeter =  \frac{1}{1000000}

As 1 with 6 zeros, we convert it into exponential form.

=  >  \frac{1}{10^{6} }

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

=  > 10^{ - 6}

Now combine with given question.

=  > 5.6 \times 10^{ - 6}

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2 years ago
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I believe that is true.
3 0
3 years ago
Read 2 more answers
According to table g, which substance forms an unsaturated solution when 80. grams of the substance are stirred into 100. grams
e-lub [12.9K]

A saturated solution is one in which no more solute is able to dissolve in a given solvent at a particular temperature. Some amount of the solute is left undissolved in the solution.

Unsaturated solution has solute in lower proportions than required to form a saturated solution.

Supersaturated solution has solute in amounts greater than a saturated solution.

We can take the help of solubility curve in order to find out the amount of a salt required to prepare a saturated solution of that salt at a particular temperature.

The solubility of KI at 10^{0}C is 136 g/ 100 mL water

The solubility of KNO_{3} at 10^{0}C is 21 g/100 mL water.

The solubility of NaNO_{3} at 10^{0}C is 80 g/100 mL water.

The solubility of NaCl at 10^{0}C is 38 g/ 100 mL water.

So the correct answer will be KI, as it would need 136 g KI / 100 mL water to form a saturated solution at 10^{0}C.So, if we have 80g KI/ 100mL water it would be an unsaturated solution.

4 0
3 years ago
Read 2 more answers
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