A bird on a long migration flies 63 kilometers per hour for 2900 kilometers. How long does he fly?

What two-step process separated the cans into aluminum, steel, and tin?

Vikentia [17]

Answer:

Magnet

Durability and heaviness.(texture)

Explanation:

Magnet can be use to separate Aluminum from mixture of steel and aluminum.

Though aluminum and steel look alike but magnet can be use to separate it.

If the can attract the magnet or magnet stick to the can, it is a steel can. Aluminum does not stick to magnet.

A mixture of Aluminum and tin can also be separated by magnet.

Tin attract magnet but tin is more durable, heavy and does not corrode easily.

When u touch the three cans, tin is heavy and durable.

4 0

2 years ago

n the diagram shown, when an object ‘X’ is brought near the ring shaped magnet, the magnet moves away from it. Four friends are

kati45 [8]

Answer:

Akash

Explanation:

it could be a magnet with the same poles facing eachoher

6 0

1 year ago

During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul

devlian [24]

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

7 0

2 years ago

Read 2 more answers

Consider the air over a city to be a box 100km on a side that reaches up to an altitude of 1.0 km. Clean air is blowing into the

Mariana [72]

Explanation:

It is known that equation for steady state concentration is as follows.

$C_{a} = \frac{QC}{Q + kV}$

where, Q = flow rate

k = rate constant

V = volume

C = concentration of the entering air

Formula for volume of the box is as follows.

V = $a^{2}h$

= $100 \times 100 \times 1$

= $10000 km^{3}$

Now, expression to determine the discharge is as follows.

Q = Av

= $100 \times 1 \times \frac{4 m}{s} \times \frac{km}{1000 m}$

= 0.4 $km^{3}/s$

And, m (loading) = 10kg/s,

k = 0.20/hr

as 1 $km^3 = 10^{12}$ L (if u want kg/L as concentration)

Now, calculate the concentration present inside as follows.

$C_{in} = \frac{10kg/s}{0.4 km^3/s}$

= 25 $kg/km^3$

Now, we will calculate the concentration present outwards as follows.

$C_{out} = {C_{in}}{(1 + k \times t)}$ ,

and, t = $\frac{V}{Q}$

= 25000 s or 6.94 hr

Hence, $C_{out} = \frac{25}{(1 + 0.20 \times 6.94)}$

= 10.47 $kg/km^3$

Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is $C_{out} = 10.47 kg/km^3$ and $C_{in} = 25 kg/km^3$ .

4 0

1 year ago

An unknown element X has the following isotopes: ²⁵X (80.5% abundant) and ²⁷X (19.5% abundant). What is the average atomic mass

IgorC [24]

Answer:

25.39

Explanation:

Given parameters:

Abundance of X-25 = 80.5%

Abundance of X - 27 = 19.5%

Unknown:

Average atomic mass of X = ?

Solution:

The average atomic mass of X can be derived using the expression below:

Average atomic mass = (abundance x mass of X - 25) + (abundance x mass of X - 27)

Average atomic mass = (80.5% x 25) + (19.5% x 27) = 25.39

5 0

1 year ago