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vladimir1956 [14]
3 years ago
5

Eriq, a chemist, is running tests with four unknown elements. He has found that they all bond the same way, so he knows that the

y are in the same group on the periodic table of elements. The table shows other properties that Eriq observed. A 4-column table with 4 rows titled properties of four unknown elements. The first column titled element number has entries 1, 2, 3, 4. The second column titled melting point has entries 266 K, 172 K, 387 K, 53 K. The third column titled boiling point has entries 332 K, 239 K, 457 K, 85 K. The fourth column titled phase at room temperature has entries liquid, gas, solid, gas. These four elements are most likely in group 15. 16. 17. 18.
Chemistry
2 answers:
Reil [10]3 years ago
4 0

Answer:

in group 17

Explanation:

just took the quiz

kicyunya [14]3 years ago
3 0

Answer:

its group 17.

Explanation:

i got 100% on the quiz

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6 0
3 years ago
1. How many molecules of H,O are in 4.32 moles?
Korvikt [17]

Answer:

dont know

Explanation:

8 0
3 years ago
In which process are simple materials chemically combined to form more complex materials?
allsm [11]

Answer:

A) synthesis

I hope this helps!

3 0
3 years ago
20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as
shepuryov [24]

The reaction forms 98.76 g AlCl_3.  

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.  

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al

Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of AlCl_3 we can obtain from each reactant.  

<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

4 0
3 years ago
Which of the following is true for the theoretical yield of a reaction? (1 point)
just olya [345]

Answer:

It is always twice the value of the actual yield.

Explanation:

3 0
2 years ago
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