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lilavasa [31]
4 years ago
8

Which of the following elements would form metallic bonds?Check all that apply.

Chemistry
1 answer:
andrew11 [14]4 years ago
4 0
All metals in the periodic table form metallic bonds. This includes every element in the periodic table except those in the non-metal group. So elements that are Polyatomic nonmetal,  Diatomic nonmetal, Noble gas are those who DON'T FORM.

Answer: Silver, Rhenium, Lithium, <span>Bismuth </span>will form metal bonds.


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How do tectonic plates move at convergent subduction boundaries? Insert a drawing with the arrows that show the movement
boyakko [2]

Answer:

If two tectonic plates collide, they form a convergent plate boundary. Usually, one of the converging plates will move beneath the other, a process known as subduction. ... As the sinking plate moves deeper into the mantle, fluids are released from the rock causing the overlying mantle to partially melt.

4 0
3 years ago
Assume that the test tube shown started out having 10.00 g of mercury(II) oxide. After heating the test tube briefly, you find 1
anyanavicka [17]

This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.

Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

m_{HgO}^{consumed}=10.00g-1.35 g=8.65 g

Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

m_{O_2}=8.65g-8.00g=0.65g

Learn more:

  • brainly.com/question/14502981
  • brainly.com/question/14236219
3 0
3 years ago
Indicate the bond polarity (show partial positive and partial negative ends) in the following bonds a) C-O b) P-H c) H-Cl d) Br-
Anarel [89]
Nun nun hmm I’m ummm I don’t wanna see how much
6 0
3 years ago
If 84.1 g of NaOH and 51.0 g of Al and 25.0 g H20 react which chemical is the limiting reactant?
12345 [234]

Answer:

  • <u><em>H₂O</em></u>

Explanation:

<u>1. Chemical quation</u>

The reaction of aluminium, sodium hydroxide and water is represented by the balanced chemical equation:

  • 2Al(s) + 2NaOH(s) + 6H₂O(l) → 2Na[Al(OH)₄] (aq) + 3H₂(g) ↑

The coefficients of each reactant and product give the theoretical mole ratios.

To find the limiting reactant you compare the theoretical ratios with the ratio of the available substaces.

<u>2. Theoretical mole ratio:</u>

  • 2 mol Al : 2 mol NaOH : 6 mol H₂O

Equivalent to

  • 1 mol Al : 1 mol NaOH : 3 mol H₂O

<u>3. Actual ratio</u>

a) Convert each mass to number of moles

Formula:

  • number of moles = mass in grams / molar mass

Al:

  • molar mass = atomic mass = 26.982g/mol
  • number of moles = 51.0g / 26.982g/mol = 1.89 mol

NaOH:

  • molar mass = 39.997g/mol
  • number of moles = 84.1g / 39.997g/mol = 2.10 mol

H₂O:

  • molar mass = 18.015g/mol
  • number of moles = 25.0g / 18.015g/mol = 1.39 mol

Divide all the mole amounts by the least number:

  • Al: 1.89/1.39 = 1.36
  • NaOH: 2.10 = 1.52
  • H₂O: 1.39 = 1.00

  • 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O

<u>4. Comparison</u>

<u />

Theoretical ratio:

  • 1 mol Al : 1 mol NaOH : 3 mol H₂O

Actual ratio:

  • 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O

     Multiply by 3:

  • 4.08 mol Al : 4.56 mol NaOH : 3.00 mol H₂O

Now, yo can see that the first two are in excess with respect the third one, making that the water consumes first, before any of the other two consumes. Therefore, the limiting reactant is the water.

8 0
3 years ago
Student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combin
nirvana33 [79]

Answer:-  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}

Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.

To calculate the heat absorbed or released we use the formula:

q=ms\Delta T

q = heat absorbed or released

m = mass of solution

s = specific heat capacity

and \Delta T = change in temperature

mass of solution = mass of solute + mass of solvent

mass of solution = 5.00 g + 100.0 g = 105.0 g

(note:- density of pure water is 1 g per mL so the mass is same as its volume)

\Delta T = 33.0 - 23.0 = 10.0 degree C

s = 4.18\frac{J}{g.^0C}

Let's plug in the values in the formula and calculate q.

q = 105.0g(4.18\frac{J}{g.^0C})(10.0^0C)

q = 4389 J

To calculate the enthalpy of solution that is \Delta H_s_o_l_n we convert q to kJ and divide by the moles of solute.

moles of solute = 5.00g(\frac{1mol}{42.39g})

= 0.118 moles

q = 4389J(\frac{1kJ}{1000J}) = 4.389 kJ

\Delta H_s_o_l_n=\frac{4.389kJ}{0.118mol}

\Delta H_s_o_l_n = 37.2\frac{kJ}{mol}

Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.

So,  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}


4 0
3 years ago
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