All categories

2 years ago

Use the drop-down menus to complete each statement.

1 answer:

8 0

Question 1: A material that causes a wave to bounce off it is called the. Answer : Reflection

Question 2: A material that takes in a wave when the wave hits is called the. Answer: Absorber

You might be interested in

A 10.0 g ice cube is placed into 250 g of water with an initial temperature of 20.0 C. If the water drops to a temperature of 16

Radda [10]

the mass of ice taken = 10 g

the mass of water = 250 g

initial temperature of water = 20 C

the final temperature of water = 16. 8 C

specific heat of water = 4.18 J/g*K

the heat absorbed by ice to melt = heat loss by water

heat loss by water = mass X specific heat of water X change in temperature

heat loss by water = 250 X 4.18 X (20-16.8) = 3344 Joules

heat gained by ice = 3344 J

heat gained by ice = enthalpy of fusion X moles of ice

moles of ice = mass / molar mass = 10 / 18 = 0.56 moles

enthalpy of fusion = 3344 / 0.56 = 5971.43 J / mole

3 0

2 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3

maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

HF = 1.62g

H₂O = 516g

F⁻ = 0.163g

H₃O⁺ = 0.110g

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

7 0

3 years ago

What tool best allows you to make observations of an objects color and texture? Balance Scale, Digital Camera, Digital Thermomet

garri49 [273]

Answer:

digital camera

Explanation:

4 0

3 years ago

How can you make a solution more saturated?

NNADVOKAT [17]

Heat the solution till all the water molecules are no longer in the solution.
Or add more solute till it stops dissolving.(maximum point of dissolvtion.)

7 0

3 years ago