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3 years ago

The density of plutonium is 19.84 g/cm². What is its mass if the container has volume of 250 cm?

Chemistry

1 answer:

serious [3.7K]3 years ago

4 0

Answer:

the answer is 7.936X 10⁻⁷kg.... the unit for volume must be cm³

Explanation:

we know,

ρ =m/v

given,

ρ =19.84/1000X(100)²= 1.984X10⁻⁶kg/m²

V (volume of the container means the volume of the element)=2.5 m³

NOW putting in the equation

1.984X10⁻⁶/2.5=m

or, m=7.936X 10⁻⁷kg(ANS)

thank u(i am not 100% sure but think it shoud be the answer)

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Fill in the coefficients that will balance the following reaction: a0NH4Cl + a1Ag3PO4 → a2AgCl + a3(NH4)3PO4

rosijanka [135]

The equation of the reaction before balancing is
a0NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
PO₄³⁻ ion is balanced.
on the left side, theres 1 (NH₄⁺) ion and right side 3 (NH₄⁺) ions. Therefore if we put the coefficient for NH₄Cl, we will obtain the following equation
3 NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
3 Ag⁺ ions on the left side and 1 Ag⁺ ion on the right side, so if we put the coefficient of AgCl as 3, following equation obtained
3 NH₄Cl + a1Ag₃PO₄ → 3 AgCl + a3(NH₄)₃PO₄
Cl⁻ ions are also balanced now, 3 on either side.
a1 and a3 are 1 as those compounds are as it is, so coefficient is 1 for both

balanced equation is as follows
3 NH₄Cl + Ag₃PO₄ → 3 AgCl + (NH₄)₃PO₄
coefficients are
a0 - 3
a1 - 1
a2 - 3
a3 - 1

8 0

3 years ago

What is the speed of the silver sphere at the moment it hits the ground? (Assume that energy is conserved during the fall and th

uranmaximum [27]

Complete question:

Consider the two spheres shown here, one made of silver and the other of aluminum. The spheres are dropped from a height of 2.1 m. (composition of sphere : density is 10.49 g/cm³, volume is 196 cm³), (composition of aluminum : density is 2.7 g/cm³, volume is 196 cm³).

What is the speed of the silver sphere at the moment it hits the ground? (Assume that energy is conserved during the fall and that 100% of the sphere’s initial potential energy is converted to kinetic energy by the time impact occurs.)

Answer:

The speed of the silver sphere at the moment it hits the ground is 6.416 m/s

Explanation:

Potential energy of the silver sphere = mgh

where;

g is acceleration due to gravity = 9.8 m/s²

h is height above the ground = 2.1m

m is mass of the silver sphere =?

Density = mass/volume

mass of the silver sphere = density X volume = (10.49 g/cm³) X (196cm³ )

mass of the silver sphere = 2056.04 g = 2.056 kg

Potential energy of the silver sphere = (2.056 X 9.8 X 2.1) J

Potential energy of the silver sphere = 42.31248 J

If energy is conserved and 100% potential energy is converted to kinetic energy, the speed of the silver sphere will be calculated as follows:

Kinetic Energy (K.E) = 1/2mv²

where;

m is mass of the sphere in kg = 2.056kg

v is the velocity of the silver sphere = ?

If energy is conserved; Potential energy = Kinetic energy

42.31248 = 1/2mv²

v² = (42.31248 X 2)/m

v² = (42.31248 X 2)/2.056

v² = 41.16

v = √41.16

v = 6.416 m/s

Therefore, the speed of the silver sphere at the moment it hits the ground is 6.416 m/s

3 0

3 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

irinina [24]

QUICK ANSWER

J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.

7 0

3 years ago

Read 2 more answers

At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of

Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.031M/atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

$C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = $0.0905mol/L$

Volume of solution = 425 mL

Putting values in above equation, we get:

$0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g$

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

8 0

3 years ago

The boiling point of water is 100.00 °C at 1 atmosphere.

Oksana_A [137]

Answer:

The solution's boiling point is 100,2°C and its molality is 0,13 m

Explanation:

This is the colligative propertie about elevation of boiling point

ΔT = Kb . m . i where

ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure

Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)

m means molality (moles of solute in 1kg of solvent)

i means theVan 't Hoff factor ( degree of dissociation for a compound)

IT HAS NO UNITS

NiI2 ---> Ni2+ + 2I- (we have 1 Ni2+ and 2 I-), the i for this, is 3

The 11,11 g of the salt are in 272,2g of water but I need to know how many mass of the salt is in 1000 g of water (1000 g is 1 kg) so the rule of three is:

272,2g ____ 11,11g

1000g _____ (1000 g . 11,11g) / 272,2g = 40,81g

As the molar mas of NiI2 is 312.5 g/mol, the moles of salt are, mass/molar mass, 40,81g /312.5 g/mol = 0,130 moles

T° of b p sl - 100°C = 0,52 °C.kg/m . 0,130 m/kg . 3

T° of boiling point solution = (0,52 °C.kg/m . 0,130 m/kg . 3) + 100°C

T° of boiling point solution = 100,2°C

4 0

4 years ago