All categories

4 years ago

Which method is appropriate for separating a mixture of iron and aluminum?

1 answer:

5 0

The method which is appropriate for separating a mixture of iron and aluminum is A. using a magnet.
This is because aluminum does not have magnetic properties (thus it will be unaffected by the magnet), so it is easy to sort out all iron from the mixture using a magnet.

You might be interested in

What is the proper way of collection, preservation and transition of hair?

Murrr4er [49]

An investigator can collect hairs they observe visually (with tweezers or by hand), and they can also use clear tape to lift non-visible hair from a variety of surfaces, such as clothing. Other methods of hair sample collection include combing and clipping methods.

7 0

3 years ago

Helppppp pls it’s due by today

Andrews [41]

Answer:

i can't see the picture

Explanation:

4 0

3 years ago

Read 2 more answers

PLEASE HELP

neonofarm [45]

Answer:

The answer to your question is 0.10 M

Explanation:

Data

Molarity = ?

mass of Sucrose = 125 g

volume = 3.5 l

Formula

Molarity = moles / volume

Process

1.- Calculate the molar mass of sucrose

C₁₂H₂₂O₁₁ = (12 x 12) + (1 x 22) + (16 x 11)

= 144 + 22 + 176

= 342 g

2.- Convert the mass of sucrose to moles

342 g of sucrose ------------------- 1 mol

125 g of sucrose -------------------- x

x = (125 x 1) / 342

x = 0.365 moles

3.- Calculate the molarity

Molarity = 0.365 / 3.5

4.- Result

Molarity = 0.10

5 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

4 years ago

You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p

Elenna [48]

Answer:

<u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

$PV=nRT$

Where:

P = pressure = 1.25 atm
V = volume = 25.2 liter
R = Universal constant of gases = 0.08206 atm-liter/K-mol
T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
n = number of moles

Solving for n:

$n=\frac{PV}{RT}$

Substituting:

$n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol$

8 0

3 years ago