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3 years ago

NaAlSi3O8 number of atoms persent

Physics

1 answer:

ad-work [718]3 years ago

5 0

Answer is in the photo. I can only upload it to a file hosting service. link below!

tinyurl.com/wpazsebu

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A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

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3 years ago

Maureen takes notes in class. Wave Interactions

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3 years ago

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3. If a voltage difference of 3.0 V causes a

MAXImum [283]

Answer:

it will be 1.8. voltage

that is wat I can't say

5 0

3 years ago

If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years

OLEGan [10]

20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.

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3 years ago

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘C and 1.00 atm. The molar mass of oxygen is 32

VashaNatasha [74]

1) 5765 mol

First of all, we need to find the volume of the gas, which corresponds to the volume of the room:

$V=7.00 m\cdot 8.00 m\cdot 2.50 m=140 m^3$

Now we can fidn the number of moles of the gas by using the ideal gas equation:

$pV=nRT$

where

$p=1.00 atm=1.01\cdot 10^5 Pa$ is the gas pressure

$V=140 m^3$ is the gas volume

n is the number of moles

R is the gas constant

$T=22.0^{\circ}+273=295 K$ is the gas temperature

Solving for n,

$n=\frac{pV}{RT}=\frac{(1.01\cdot 10^5 Pa)(140 m^3)}{(8.314 J/molK)(295 K)}=5765 mol$

2) 184 kg

The mass of one mole is equal to the molar mass of the oxygen:

$M_m = 32.0 g/mol$

so if we have n moles, the mass of the n moles will be given by

$m : n = 32.0 g/mol : 1 mol$

since n = 5765 mol, we find

$m=\frac{(5765 mol)(32.0 g/mol)}{1 mol}=1.84\cdot 10^5 g=184 kg$

3 0

3 years ago

NaAlSi3O8 number of atoms persent​

NaAlSi3O8 number of atoms persent