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3 years ago

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If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? ball A,

ball B, Their densities are equal.
If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density? ball C, ball D, Their densities are equal.

If two balls have the same mass, but ball P is twice as large as ball Q, which one will have the greater density? ball P, ball Q, Their densities are equal.

If ball X is twice as big as ball Y and weighs only half as much as ball Y, then which one will have the greater density? ball X, ball Y, Their densities are the same.

2 answers:

castortr0y [4]3 years ago

7 0

Answer:

The correct ansers are

Slot 1 is Ball A

Slot 2 is there deseties are equal

Slot 3 is Ball Q

Slot 4 is Ball Y

Explanation:

Took the asingment 2020 Edgeunuty

PLZ GIVE BRAINLIESt

rewona [7]3 years ago

5 0

Density (D) is defined as the amount or mass (m) of a substance present in a unit volume(V). It can be expressed mathematically as:

Density = Mass/Volume

i.e. D = m/V -------(1)

Units: g/cm3

a)

If m1, V1 and D1 = mass, volume and density respectively of ball A

m2, V2 and D2 = mass, volume and density respectively of ball B

It is given that: V1 = V2; but m1 = 2m2

Based on equation (1) we have:

D1/D2 = (m1/V1)* (V2/m2) = (2m2/V2)*(V2/m2) = 2

Thus, density of ball A is twice that of B.

Ans: Ball A will have a greater density than B

b)

If m1, V1 and D1 = mass, volume and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

It is given that:

V1 = 3V2 and m2 = 1/3(m1) i.e m1 = 3m2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (3m2/3V2)*(V2/m2) = 1

Thus, D1 = D2

Ans: Their densities are equal

c)

If m1, V1 and D1 = mass, volume and density respectively of ball P

m2, V2 and D2 = mass, volume and density respectively of ball Q

It is given that:

m1 = m2 but V1 = 2V2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (m2/2V2)*(V2/m2) = 1/2

Thus, D1 = 1/2(D2)

Ans: Ball Q will have a greater density.

d)

If m1, V1 and D1 = mass, volume and density respectively of ball X

m2, V2 and D2 = mass, volume and density respectively of ball Y

It is given that:

V1 = 2V2 and m1 = 1/2(m2)

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = ((1/2(m2)/2V2)*(V2/m2) = 1/4

Thus, D1 = 1/4(D2)

Ans: Ball Y will have a greater density.

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damaskus [11]

I would always start by balancing your carbons, and then balancing the rest from there.

1. C2H5OH + O2 —> CO2 + H2O - You have two carbons on the left and one on the right. Multiply CO2 by 2.
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Now balance hydrogen. You have 6 on the left and 2 on the right. Multiply H2O by 3.
C2H5OH + O2 —> 2CO2 + 3H2O

Now balance oxygen. You have 3 on the left and 7 on the right. You need 4 more on the left. Don’t multiply the C2H5OH by anything because that will change the numbers of everything else too. Multiply O2 by 3 instead.
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Check that all atoms are now balanced, and you’re good.

2. Same process as before.

First carbons - C3H8 + O2 —> 3CO2 + H2O
Then hydrogens - C3H8 + O2 —> 3CO2 + 4H2O
Then oxygens - C3H8 + 5O2 —> 3CO2 + 4H2O

3. Same again.

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2 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

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