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3 years ago

How many grams of oxygen (O2), x, are needed to react with the 700 g of iron to produce 1000 g of iron (III) oxide, Fe2O3?

1 answer:

6 0

As we know that, molecular mass of ferric oxide, Fe2O3, is 159.69 grams.
Out of which, iron contributes 111.69 g (2 X 55.845 g) and oxygen contributes
48 g (3 X 16 g).
Each gram of iron (III) oxide contains 111.69/159.69 g of iron and 48/159.69
g of oxygen.
To produce 1000 g iron (III) oxide we need,
Iron = 111.69*1000/159.69 = 699.42 g
Oxygen = 48*1000/159.69 = 300.58 g

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A train crosses a bridge that is 1300 m long in 1 minute and 5 seconds.

Nat2105 [25]

Answer:

I'm pretty sure it's 20m/s because 1300m divided by 65 seconds is 20 so I think it's 20m/s

Explanation:

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3 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

6 0

3 years ago

Read 2 more answers

Infrared light of wavelength 2.5 µm illuminates a 0.20-mm-diameter hole. What is the angle of the first dark fringe in radians?

stepan [7]

Answer:

Θ=0.01525 rad

Θ=0.87°

Explanation:

Given data

wavelength λ=2.5 µm =2.5×10⁻⁶m

Diameter d=0.20 mm =0.20×10⁻³m

To find

Angle Θ in radians and degree

Solution

Circular apertures have first dark fringe at

Θ=(1.22λ)/d

Substitute the given values

Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m

Θ=0.01525 rad

Θ=0.87°

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3 years ago

Which atom is most likely to accept the electrons to form from an ionic bond

Mekhanik [1.2K]

a nonmetal with the amount of valence electrons that will complete atom

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3 years ago

The legend that Benjamin Franklin flew a kite as a storm approached is only a legend; he was neither stupid nor suicidal. Suppos

UNO [17]

Answer:

The current on the water layer = 1.64×10^-3A

Explanation:

Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:

I = V × pi[(Rwater + Rstring)^2 - (Rstring)^2/ ( Resitivity × L)

I =[ 166×10^6 ×3.142[(0.519×10^-4) + (2.15×10^-3])^2 - ( 2.15×10^-3)^2] / ( 183 × 831)

I =[ 521572000(4.848×10^6)- 4.623×10^-6]/ 154566

I = 252.83 -(4.623×10^-6)/ 154566

I = 252.83/154566

I = 1.64× 10^-3A

3 0

3 years ago