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3 years ago

What is the mass of a cylinder of lead that is 1.80 in in diameter, and 4.12 in long. the density of lead is 11.4 g/ml?

Physics

2 answers:

AlekseyPX3 years ago

7 0

Answer:The mass of the lead cylinder is 1,995.898 g.

Explanation:

Radius of the cylinder,r = $\frac{Diameter}{2}=\frac{1.80 inches}{2}=0.9 inches$

Height of the cylinder,h = 4.12 inches

Volume of the cylinder: $\pi r^2h=3.14\times (0.9 inches)^2\times 4.12 inches=10.47 inches^3$

$inches^3=16.3871 ml$

Volume of the cylinder in ml= $10.47\times 16.3871 ml=171.57 ml$

Density of the lead cylinder = 11.4 g/ml

$density=\frac{Mass}{Volume}$

$Mass=11.4 g/ml\times 171.57 ml=1,955.898 g$

The mass of the lead cylinder is 1,995.898 g.

Sedbober [7]3 years ago

5 0

1950 g This is the answer due to the kilograms of lead being distributed

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You treat 9.540 g of the mixture with the acid and isolate 9.355 g of nacl. what is the weight percent of each substance in the

Mila [183]

The weight percentage of sodium carbonate in the mixture is = 67.71%

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

<h3>What does "molar mass" ?</h3>

The total mass throughout grams of all the atoms needed to form a molecule per mole is what makes up the molar mass, also known as the molecular weight. Grams per mole is the unit measuring molar mass.

<h3>According to the given information:</h3>

The interaction between sodium carbonate and hydrochloric acid has the following equation:

Na₂CO₃ + HCl --> NaCl + CO₂ + H₂O

The reaction's balanced equation is,

Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O -------(2).

When sodium hydrogen carbonate as well as hydrochloric acid react, the following equation results.

NaCO₃ + HCl --> NaCl + CO₂ + H₂O ----------------(3)

The Molar mass :

Na₂CO₃ = 106g/mol

NaCO₃ = 84g/mol

NaCl = 58.5g/mol

The mass of the mixture is Na₂CO₃/NaCO₃ = 9.540g

The molar masses of the constituent elements that make up the compounds are added to determine the molar weight of the substances. In both chemical equations (2) and (3), the unitary technique is employed to calculate the mass of the mixture based on the number of moles (3).

The mass of the sodium carbonate and sodium bicarbonate is calculated with the help of the mass of NaCl formed by the mixture of Na₂CO₃/NaCO₃ with HCl.

The mixture has the following percentage of sodium carbonate:

% of Na₂CO₃ = $\frac{x \times 106}{9.540} \times 100\\$

$=\frac{0.061 \times 106}{9.540} \times 100$

= (6.46/9.540)*100

= 67.71%

The weight percentage of sodium carbonate in the mixture is = 67.71%

The mixture has the following percentage of NaHCO3:

% of NaHCO₃ = $\frac{y \times 84}{9.540} \times 100$

= $=\frac{0.037 \times 84}{9.540} \times 100$

= (3.108/9.540)*100

= 32.57%

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

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I understand that the question you are looking for is:

A mixture of sodium carbonate and sodium hydrogen carbonate is treated with aqueous hydrochloric acid. the unbalanced equations for the resulting's reactions are:

Na2CO3 (s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

NaHCO3(s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

You treat 9.540g of Na2CO3/NaHCO3 mixture with an excess of aqueous HCl and isolate 9.355g of NaCl. What is the weight percent of each substance in the mixture?

5 0

1 year ago

A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m

Feliz [49]

Explanation:

Let the speeds of father and son are $v_f\ and\ v_s$ . The kinetic energies of father and son are $K_f\ and\ K_s$ . The mass of father and son are $m_f\ and\ m_s$

(a) According to given conditions, $K_f=\dfrac{1}{3}K_s$

And $m_s=\dfrac{1}{4}m_f$

Kinetic energy of father is given by :

$K_f=\dfrac{1}{2}m_fv_f^2$ .............(1)

Kinetic energy of son is given by :

$K_s=\dfrac{1}{2}m_sv_s^2$ ...........(2)

From equation (1), (2) we get :

$\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}$ ..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

$\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}$

$v_s^2=4(v_f+1.5)^2$

Using equation (3) in above equation, we get :

$v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s$

(b) Put the value of $v_f$ in equation (3) as :

$v_s=7.09\ m/s$

Hence, this is the required solution.

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4 years ago

2. A can filled with sand has a mass of 0.65kg is swung overhead in a horizontal circle of radius 0.70m at a constant rate of 2.

Aliun [14]

<h3><u>Answer</u>;</h3>

≈ 5 Kgm²/sec

<h3><u>Explanation</u>;</h3>

Angular momentum is given by the formula

L = Iω, where I is the moment of inertia and ω is the angular speed.

I = mr², where m is the mass and r is the radius

= 0.65 × 0.7²

= 0.3185

Angular speed, ω = v/r

= (2 × 3.142 × r × 2.5) r

= 15.71

Therefore;

Angular momentum = Iω

= 0.3185 × 15.71

= 5.003635

<u>≈ 5 Kgm²/sec</u>

6 0

3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

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4 0

3 years ago

Difference between on pitch and frequency

shepuryov [24]

Answer:

A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. I hope I got it correct !!

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