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vaieri [72.5K]
3 years ago
9

Suppose a star the size of our Sun, but of mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 17 days. If

it were to undergo gravitational collapse to a neutron star of radius 15 km, losing 3/4 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times
Physics
1 answer:
Tanya [424]3 years ago
4 0
Use the conservation of angular momentum; angular momentum at the beginning = angular momentum at the end 
Conservation of angular momentum: 
I1 w1 = I2 w2 
Where I is the moment of inertia. For a sphere, I=2/5 m R^2. Substituting into the equation above we get 
w2 = I1 w1 / I2 = w1 m1 R1^2 / (m2 R2^2) 
w2 = w1 4 * (R1/R2)^2
= 4*(1)*(7E5/7.5)^2
= 3.48E10 revs/(17days)
= 2.04705882 x 10^9 revs/sec
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calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
What was one main point of Dalton's atomic theory
goblinko [34]
All matter is composed of atoms, indestructible building blocks.
4 0
3 years ago
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A student measures the mass 8cm block of brown sugar to be 12.9g. what is the density of the brown sugar.
antoniya [11.8K]

Correction

A student measures the mass <em><u>8cm3</u></em> block of brown sugar to be 12.9g. what is the density of the brown sugar

Answer:

1.6\ g/cm^{3}

Explanation:

Density is defined as mass per unit volume of an object expressed as \rho=\frac {m}{v} where \rho is the density, m is the mass of sugar and v is the volume of the sugar. Considering that the volume is given as 8cm3 for sugar then we substitute this for v and mass of 12.9 g we substitute for g then the density will be

\rho=\frac {12.9 g}{8 cm3}=1.6125\ g/cm^{3}\approx 1.6\ g/cm^{3}

8 0
3 years ago
Which type of rock can only form below earths surface?
Inessa [10]

Answer: Igneous it forms because of magma but magma is under the earths surface so its Igneous

Explanation:

7 0
2 years ago
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Strontium−90 is one of the products of the fission of uranium−235. This strontium isotope is radioactive, with a half-life of 28
Ludmilka [50]

Answer : The time passed in years is 20.7 years.

Explanation :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{28.1\text{ years}}

k=2.47\times 10^{-2}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.47\times 10^{-2}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant  = 1.00 g

a - x = amount left after decay process = 0.600 g

Now put all the given values in above equation, we get

t=\frac{2.303}{2.47\times 10^{-2}}\log\frac{1.00}{0.600}

t=20.7\text{ years}

Therefore, the time passed in years is 20.7 years.

8 0
3 years ago
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