Atoms of which of elements will form covalent bonds in a compound

A burner is placed below the right side of a beaker. Which image shows the motion of the liquid that will result?

matrenka [14]

Answer:

the result will be the whole surface of the liquid boilling

Explanation:

3 0

2 years ago

I need the answers of these

Anni [7]

Answer:

Hope this helps :D

Explanation:

Metal: Aluminum and Copper

Non-Metal: Hydrogen and Flourine

Acid: Sulfuric Acid and Phosphoric Acid

Alkali Metals: Hydrogen and Lithium

Compounds: Water and Carbon Dioxide

Elements: Carbon and Oxygen

4 0

1 year ago

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

GalinKa [24]

<h2>a) The rate at which $NO_2$ is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen $O_2$ is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s

Rate in terms of disappearance of $O_2$ = $-\frac{1d[O_2]}{dt}$

Rate in terms of appearance of $NO_2$ = $\frac{1d[NO_2]}{2dt}$

1. The rate of formation of $NO_2$

$-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}$

$\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s$

2. The rate of disappearance of $O_2$

$-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}$

$-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

Learn more about rate law

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https://brainly.in/question/1297322

7 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

3 years ago

Consider this equilibrium reaction between carbon monoxide and hydrogen gas, occurring in a sealed flexible container. CO(g) + 3

wariber [46]

Answer:

More H2(g) is added to the container : Towards products.

CO is removed from the container : Towards reactants.

More CH4(g) is added to the container : Towards reactants

H2O(g) is removed from the container : Towards products.

The contents of the container are heated up. : Towards the reactants.

The contents of the container are cooled down : Towards the products.

The pressure inside the container is increased. :Towards the products

The container is stretched to increase the volume: Towards the reactants.

Explanation: :

CO(g) + 3 H2g) → CH4(g) + H2O(g)+ heat

There is released heat, so this reaction is exothermic

If the H2 concentration is increased, the system will try to change the concentration change by shifting the balance to the right, and thus the concentration of products will increase. Towards products.

If the CO is removed, the system will try to change this situation by shifting the balance to the left, and thus the concentration of reactants will increase, the concentration of products will decrease. Towards reactants.

If the CH4 concentration is increased, the system will try to change the concentration change by shifting the balance to the left, and thus the concentration of reactants will increase. Towards reactants

If the H2O is removed, the system will try to change this situation by shifting the balance to the right, and thus the concentration of products will increase, the concentration of products will decrease. Towards products.

If the temperature is increased, the system will reduce the amount of heat released. So the balance will shift to the left. Towards the reactants.

This because the extra heat / energy must be used.

If the temperature is decreased, the system will produce more heat So the balance will shift to the right. Towards the products.

This because more heat /energy needs to be produced to make up for the loss of heat (energy).

If the pressure is increased, the system will shift to the side with fewer moles of gas. In this case, there are 4 moles on the left and 2 moles on the right. So the balance will shift to the right. Towards the products. An increase of pressure has the same effect on the equilibrium as a decrease of the volume.

If the volume is increased, this means the pressure is decreased, the system will shift to the side with most moles of gas. In this case, there are 4 moles on the left and 2 moles on the right. So the balance will shift to the left. Towards the reactants. An increase of volume has the same effect on the equilibrium as a decrease of the pressure.

6 0

3 years ago