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mamaluj [8]
3 years ago
12

A scientist is conducting an experiment with sand. Some iron filings accidentally fell into the sand. What could the scientist u

se to separate out the two substances? 25 points!! PLZZ HELP
Chemistry
2 answers:
anzhelika [568]3 years ago
7 0
Get a magnet in a bag (inside out) and move the magnet all over the sand/iron and you can collect all the iron and turn the bag right side out and zip up the bag. Now you have the sand and iron separated and it will be harder to accidentally mix them later.
lara31 [8.8K]3 years ago
7 0
Get a magnet and hold the magnet right above the sand so the iron fillings deprecate by attraction to the magnet
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What is the molar mass of magnesium chlorite (Mg(CIO2)2)?
sveticcg [70]
Hope this helps! 159.2086<span>Have a Good Day!</span>
4 0
3 years ago
What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, c2h6? in this reaction,
Vitek1552 [10]

Balance equation for combustion of ethane will be:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

To balance the equation:

1. Balance the number of carbon atom on both side:

C₂H₆(g) + O₂(g)--------> CO₂(g) + H₂O(g)

1. balance the number of carbon on both side, as in reactant there are 2 but in product one,

so , multiply the CO₂, by 2 in the product.

2. Balance the number of hydrogen on both side as in reactant the number of hydrogen is 3 but in product it is 6 so, multiplythe number of  H₂O by 3,

so multiply the number of  H₂O by 3 in product.

3. Balance the number of oxygen on both side , as 1 and 2 step increases the number of oxygen and it becomes 7 , so to balance the number of oxygen on both side by mutiplying the  number of  O₂ by 7/2 in reactant .

4. Now, doubling the equation will give balance equation that is:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

8 0
3 years ago
A 0.590 gram sample of a metal, M, reacts completely with sulfuric acid according to:M(s) +H2SO4(aq) --&gt; MSO4(aq) +H2(g)A vol
photoshop1234 [79]

Answer:

MM = 58.41 g

Explanation:

First, the data we have is according to the hydrogen which is exerting pressure. To solve this, we need to use the ideal gas equation:

PV = nRT (1)

the molar mass of any compound is calculated like this:

MM = m/n (2)

So, from (1) we solve for the moles (n) and then, this value is replace in (2).

However, before we do all that, we need to gather all the correct data.

All the species in the reaction are solid or aqueous state, with the exception of hydrogen, which is gaseous. Hydrogen is collected over water, therefore, is exerting some pressure too. The problem is not indicating if the acid or any other species is exerting pressure, so we will assume that only hydrogen and water are exerting pressure.

The total pressure exerted by the system would be:

P = Pw + PH2 (3)

We already know the total pressure which is 756 torr.

This experiment is taking place at 25 °C (298.15 K), and at this temperature, we have a reported value for water pressure which is 23.8 Torr.

Let's solve for PH2:

PH2 = P - Pw

PH2 = 756 - 23 = 733 Torr

Now, with this value, and the volume and temperature, we can calculate the moles of H2:

n = PV/RT

But first, let's convert the pressure to atm:

PH2 = 733 Torr / 760 torr * 1 atm = 0.9644 atm

now, solving for n:

n = 0.9644 * (0.255) / 0.082 * 298.15

n = 0.0101 moles

Now that we have the moles, we know that the metal and the hydrogen has a mole ratio of 1:1 according to the reaction, so, this means that:

moles M = moles H2 = 0.0101 moles

We have the moles of the metal and the mass, we can calculate the molar mass using expression (2):

MM = 0.590/0.0101

MM = 58.41 g/mol

This is the molar mass of the metal

8 0
3 years ago
A factor that does no change in an experiment is the.
kvasek [131]

Answer:

A. Controlled variable

Explanation:

a controlled variable or a constant variable is a variable that doesnt change during an experiment

5 0
3 years ago
An object was measured by a worker as
Alexeev081 [22]

Answer:.633

Explanation:

I have know idea but it was right

3 0
3 years ago
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