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3 years ago

49.2 m = _________ hm

Chemistry

1 answer:

dalvyx [7]3 years ago

4 0

The answer is 0.492 because there are 100 meters in a hectometer, and 49.2/100 is 0.492

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Which coefficients will balance the following reaction:

valentina_108 [34]

Answer:

C ) 2, 1, 2

Explanation:

The given reaction is synthesis reaction in which lithium and bromine react to form lithium bromide.

Chemical equation:

Li + Br₂ → LiBr

Balanced chemical equation:

2Li + Br₂ → 2LiBr

Step 1:

Li + Br₂ → LiBr

left hand side Right hand side

Li = 1 Li = 1

Br = 2 Br = 1

Step 2:

Li + Br₂ → 2LiBr

left hand side Right hand side

Li = 1 Li = 2

Br = 2 Br = 2

Step 3:

2Li + Br₂ → 2LiBr

left hand side Right hand side

Li = 2 Li = 2

Br = 2 Br = 2

3 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

What is the [ H3O+] of a solution with pH = (8.50x10^0)? enter your answer in scientific notation and with three sig figs.

erastovalidia [21]

Answer:

[H₃O⁺] = 3.162 × 10⁻⁹ moles / liter

Explanation:

The pH is calculated with the formula:

pH = -lg[H₃O⁺]

From here the concentration of hydronium ion (H₃O⁺) will be:

[H₃O⁺] = $10^{-pH}$

[H₃O⁺] = $10^{-8.5}$

[H₃O⁺] = 3.162 × 10⁻⁹ moles / liter

3 0

3 years ago

Read 2 more answers

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

Answer:

For (a): The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

For (b): The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

For (c): The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

For (d): The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

For (e): The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago

What happens to the energy that is lost when water freezes?

SVEN [57.7K]

As we know through the principle of conservation of energy, energy can neither be created nor destroyed. Therefore, the energy removed from the water in order to make it freeze is absorbed by the surroundings. This is why the surroundings in which freezing is taking place are below freezing. This is more easily illustrated in the example of condensation. If you were to hold a plate over a pot of boiling water, some of the water would give its energy to the plate and condense on its surface.

4 0

3 years ago

Read 2 more answers