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3 years ago

1. Put the layers of Earth in order from densest to least dense.

Chemistry

1 answer:

icang [17]3 years ago

7 0

Answer:

1-A

2-B

Explanation:

2- mantle is liquid and moves crust

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Does anyone know the anserw the symbol of the element

Komok [63]

Cesium.

Groups are the vertical columns that run up and down while periods are the horizontal rows. So to find the answer to this, go to the first column (Group 1) and find the sixth period (row 6) which will land you on Cesium, element 55.

Hope this helps!

3 0

3 years ago

Why is the melting of ice a physical change? It changes the chemical composition of water. It does not change the chemical compo

snow_lady [41]

It does not change the chemical in the composition of water. hope it helps

3 0

3 years ago

Read 2 more answers

A chemist must prepare of hydrochloric acid solution with a pH of at . He will do this in three steps: Fill a volumetric flask a

Eva8 [605]

Answer:

1.7 mL

Explanation:

<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>

Step 1: Calculate [H⁺] in the dilute solution

We will use the following expresion.

pH = -log [H⁺]

[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M

Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.

Step 2: Calculate the volume of the concentrated HCl solution

We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL

3 0

2 years ago

What is the mass of 3.0 x 1023 atoms of neon

Xelga [282]

1 mole =6.0 * 10^23 atoms
3.0*10^23 atoms = 0.5 moles
the molar mass of neon is 20.18g/mole
0.5 moles = 10.09 grams

8 0

2 years ago

Read 2 more answers

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago