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UkoKoshka [18]
4 years ago
7

Suppose you are measuring the mass of a sample using a balance that employs a standard mass of density dw = 8.0 g/ml. what is th

e buoyancy correction (m/m\'), if the sample density is less than 8.0 g/ml?
Chemistry
1 answer:
sleet_krkn [62]4 years ago
5 0

The buoyant force formula is:

B = \rho Vg

where B is Buoyant force, \rho is density, V is displaced body volume and g is 9.806 ms^{-2} (standard gravity).

The standard mass density of the sample, \rho _{1} = 8 g/ml  

If the dansity of the sample, \rho _{2} < 8 g/ml

The buoyancy correction is determined by the ratio:

\frac{B_{1}}{B_{2}} = \frac{\rho_{1} Vg}{\rho_{2} Vg}

\frac{B_{1}}{B_{2}} = \frac{\rho_{1}}{\rho_{2}}

\frac{B_{1}}{B_{2}} = \frac{\rho_{1}}{\rho_{2}}>1

Hence, the the buoyancy correction is positive.

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3 years ago
Na2CO3(aq) + CaCl2(aq) — 2 NaCl(aq) + CaCO3(s)
attashe74 [19]

Answer:

100 mL

Explanation:

Given data:

Mass of CaCO₃ produced = 2.00 g

Molarity of CaCl₂ = 0.200 M

Volume of CaCl₂ needed = ?

Solution:

Chemical equation:

Na₂CO₃  + CaCl₂    →      2NaCl + CaCO₃

First of all we will calculate the number of moles of CaCO₃.

Number of moles = mass/molar mass

Number of moles = 2.00 g / 100.09 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of CaCO₃ and CaCl₂.

              CaCO₃          :            CaCl₂

                   1                :               1

                 0.02           :              0.02

Thus, 0.02 moles of CaCl₂ react,

Volume of CaCl₂ reacted:

Molarity = number of moles / volume in L

0.200 M = 0.02 mol / volume in L

Volume in L = 0.02 mol / 0.200 M

Volume in L = 0.1 L

Volume in mL:

0.1 L × 1000 mL/1L

100 mL

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3 years ago
Of the following equilibria, only ________ will shift to the right in response to a decrease in volume. N2 (g) + 3H2 (g) 2NH3 (g
igomit [66]

Answer:

Of the following equilibria, only one will shift to the right in response to a decrease in volume.

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Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Decrease the volume

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N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

2 Fe_2O_3 (s) \rightleftharpoons 4 Fe (s) + 3O_2 (g)

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2 SO_3 (g) \rightleftharpoons 2 SO_2 (g) + O_2 (g)

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2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

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Answer:

There are 8 valence electrons

Explanation:

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