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Lady_Fox [76]
3 years ago
12

Calculate the Density of the substances below to determine if they will float or sink in water.

Chemistry
1 answer:
Reil [10]3 years ago
7 0

Explanation:

If the density of the object is more than that of water, it will sink. Otherwise it will float. The density of water is 1 g/mL.

Substance 1,

Mass, m = 450 g, Volume, V = 90 mL

Density = mass/volume

So,

d_1=\dfrac{450}{90}\\\\=5\ g/mL

It will sink.

Substance 2,

Mass, m = 35 g, Volume, V = 70 mL

Density = mass/volume

So,

d_2=\dfrac{35}{70}\\\\=0.5\ g/mL

It will float.

Substance 3,

Mass, m = 24 g, Volume, V = 12 mL

Density = mass/volume

So,

d_3=\dfrac{24}{12}\\\\=2\ g/mL

It will sink.

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Misha Larkins [42]
The stock solution contains 10.5 moles of HCl per litre. A 5.5 litre solution of 2.5M HCl contains 5.5x2.5 = 13.75moles of HCl. Since every litre of stock solution provides 10.5M HCl, the amount of stock solution needed is 13.75/10.5 = 1.309L. Therefore you would dilute 1.309L of stock solution to 5.5L
5 0
4 years ago
"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum
UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = 25^{o} C

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = C \times 0.082 \times 298.15 K


               C = 0.0285

This also means that,

  \frac{\text{moles}}{\text{volume (in L)}} = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = 2.85 \times 10^{-3
}

Now, let us assume that mass of C_{12}H_{23}O_{5}N = x grams

And, mass of C_{12}H{22}O_{11} = (1.00 - x)

So, moles of C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}

                              = \frac{x}{369}

Now, moles of C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}

                   = \frac{x}{369} + \frac{(1.00 - x)}{342}

                  = 2.85 \times 10^{-3}

             = x = 0.346

Therefore, we can conclude that amount of C_{12}H_{23}O_{5}N present is 0.346 g  and amount of C_{12}H_{22}O_{11} present is (1 - 0.346) g = 0.654 g.

4 0
3 years ago
What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B
natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

A = A_{0}e^{-kt}

where A(0) = initial concentration of the reactant

A = concentration after time 't'

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Taking ln on both sides gives:

ln[A] = ln[A]_{0}-kt

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

7 0
4 years ago
What elements can join wit other elements o form a covalent bond
Bas_tet [7]

Answer:

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Explanation:

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It is formed by the sharing of electron pair between bonded atoms.  

The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

Both atoms bonded through covalent bond.

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source: wikapedia
4 0
3 years ago
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