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hoa [83]
3 years ago
6

A weight of 22.25g was obtained when 10-mL of water at 27C was pipet to empty Erlenmeyer flask. What is the volume delivered by

the pipet?
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

the volume delivered by the pipette = 22.32 mL

Explanation:

To calculate this, let us first note that the density of water relates it weight and its volume (density = mass ÷ volume), hence we are going to use density to determine the volume.

Density of water = mass/volume = 0.997 g/ mL

mass = 22.25g

Density = 0.997g/mL

volume = ?

density = \frac{mass}{volume}\\\therefore volume = \frac{mass}{density}\\volume = \frac{22.25}{0.997}\\volume =  22.32\ mL

∴ the volume delivered by the pipette = 22.32 mL

<em>Please note that this calculation is based on the fact that the weight of the empty flask has been determined and canceled out.</em>

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Mass can be created and destroyed.
Musya8 [376]

Answer:

Conservation of Energy and Mass

The law of conservation of mass states that in a chemical reaction mass is neither created nor destroyed. ... Similarly, the law of conservation of energy states that the amount of energy is neither created nor destroyed.

5 0
3 years ago
If 0.2 g of nitrobenzene are added to 10.9 g of naphthalene, calculate the molality of the solution. (given: molar mass of nitro
In-s [12.5K]

Molality is defined as 1 mole of a solute in 1 kg of solvent.  

Molality=

\frac{Number of moles of solute}{Mass of solvent in kg}

Number of moles of solute, n=  

\frac{Given mass of the substance}{Molar mass of the substance}

Given mass of the nitrobenzene=0.2 g

Molar mass of the substance= 123.06 g mol⁻¹

Number of moles of nitrobenzene,  

n= \frac{0.2 }{123.06}

Number of moles of nitrobenzene, n= 0.0016  mol

Mass of 10.9 g of naphthalene in kg=0.0109  

Molality= \frac{0.0016}{0.0109 }

Molality= 0.146 m

7 0
3 years ago
All of the following show a periodic pattern except
AlekseyPX
<span>A. the dog ate everyone of his play toys</span>
4 0
3 years ago
sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction
prisoha [69]

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

                         C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹:       0         0         -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

\text{The relationship between $\Delta G^{\circ}$ and K  is}\\\Delta G^{\circ} = -RT \ln K

T = (25.0 + 273.15) K = 298.15 K

\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}

3 0
3 years ago
8. Which gas contributes the largest part of air? *
timurjin [86]

Answer:

Nitrogen

Explanation:

Nitrogen present 78% in the earth's atmosphere

5 0
3 years ago
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