Ask question

Ask question

All categories

2 years ago

15

A lead ball is added to a graduated cylinder containing 41.7 ml of water, causing the level of the water to increase to 96.0 mL.

What is the volume in milliliters of the lead ball, Vieras? Viead =

1 answer:

rjkz [21]2 years ago

5 0

Answer:

$Vlead ball=54.3 mL$

Explanation:

The graduated cylinder contains $41.7mL$ of water

mL is a volume unit.

Water volume = 41.7 mL

The lead ball caused an increase of volume from 41.7 mL to 96.0 mL

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

$96.0mL=Vleadball +41.7mL$

$Vlead ball=96.0mL-41.7mL$

$Vleadball=54.3mL$

This is actually true if we suppose that the lead ball is fully sunken in the water.

We always must consider that the volume difference is the volume that the sunken object is occupying in the water.

You might be interested in

The initial temperature of the water in a constant-pressure calorimeter is 24°C. A reaction takes place in the calorimeter, and

kupik [55]

Answer:

Explanation:

For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..

8 0

2 years ago

Read 2 more answers

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle

Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0

2 years ago

An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The

GuDViN [60]

Answer:Mass of Potassium chloride =1.762g

Explanation:

Mass of empty beaker = 23.100 g

Mass of beaker with Potassium chloride = 24.862g

Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride - Mass of empty beaker = 24.862-23.100 = 1.762g

8 0

2 years ago

Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. which con

enyata [817]

Answer:

The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.

Explanation:

From the question,

Container A contains 1000 atoms

Container B contains 500 atoms

<u>The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.</u>

The reason for such is due to the difference in the concentration of the isotopes. Container A which contains higher number of atoms will have the more changes of the release of the neutron as the changes of the hitting and splitting increases as the density of the atoms increases.

<u>Thus, the atoms in the container A will therefore decay faster than the atoms in the container B. </u>

8 0

2 years ago