- A 300g sample of CaCO3 was heated until 10 L of CO2 was collected

Chemistry

1 answer:

elixir [45]2 years ago

6 0

Answer:

C.12.3%. you need to use pv=nRT (ideal gas law)

You might be interested in

list down some examples of solutions that we need to prepare/make in the form of unsaturated and saturated solution

Vladimir79 [104]

Answer:

1)Carbonated water is saturated with carbon, hence it gives off carbon through bubbles.

2)Adding sugar to water until it no longer dissolves creates a saturated solution.

3)Continuing to dissolve salt in water until it will no longer dissolve creates a saturated solution.

An unsaturated tea and sugar solution would be one into which you could add more sugar and have the sugar still dissolve

5 0

2 years ago

as a 15.1-gram sample of a metal absorbs 48.75 j of heat its temperature increases 25.0 k what is the specific heat capacity of

Leviafan [203]

Specific heat capacity (c) of a material is related to the Energy Absorbed (Q), mass of the material (m) and the change in temperature (T) by the following equation:

$c= \frac{Q}{mT}$

Substituting the values of Q, m and T in the above equation, we get:

$c= \frac{48.75}{15.1*25}=0.129$

So the specific heat capacity of the metal with given conditions will be 0.129 J/g.K

4 0

2 years ago

Ammonia has a Kb of 1.8 × 10−5. Find [H3O+], [OH−], pH, and pOH for a 0.310 M ammonia solution.

lawyer [7]

Answer:

[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;

pH = 11.4; pOH = 2.6

Explanation:

The chemical equation is

$\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \,\text{OH}$^{-}$$

For simplicity, let's re-write this as

$\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$$

1. Calculate [OH]⁻

(a) Set up an ICE table.

B + H₂O ⇌ BH⁺ + OH⁻

0.310 0 0

-x +x +x

0.310-x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}$