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son4ous [18]
3 years ago
12

What two conditions must satisfy to be called matter ?

Chemistry
1 answer:
prohojiy [21]3 years ago
6 0
I can tell you one. Must have particles
You might be interested in
Question 6 *
ycow [4]

Answer:

33 g

Explanation:

Given data:

Number of moles = 5 mol

Mass of compound = 165 g

Molar mass or gram formula mass = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Now we will put the values in formula.

5 mol = 165 g/ gram formula mass

Gram formula mass = 165 g / 5 mol

Gram formula mass = 33 g/mol

7 0
2 years ago
Which sequence contains elements listed from most reactive to least reactive? *
Helga [31]

Answer:

Reactivity Series

Explanation:

Potassium

Sodium

Lithium

Calcium

Magnesium

Aluminium

Zinc

Iron

Copper

Silver

Gold

7 0
2 years ago
Each orbit surrounding an atom is allowed _____.
Juli2301 [7.4K]
Two electrons is your answer glad to help!
4 0
3 years ago
Read 2 more answers
3. During an experiment where 50.0 mL of a 1.0 M acid solution was mixed with 50.0 mL of a 1.0 M base solution, the temperature
Aleksandr-060686 [28]

Answer:

\large \boxed{\text{-61 kJ$\cdot$mol$^{-1}$}}  

Explanation:

Data:

                H₃O⁺ +  OH⁻ ⟶ 2H₂O

    V/mL:  50.0     50.0

c/mol·L⁻¹:   1.0        1.0

     

    ΔT = 6.5 °C  

      ρ = 1.210 g/mL

      C = 4.18 J·°C⁻¹g⁻¹

C_cal = 12.0 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

\text{Mass of solution} = \text{100.0 mL} \times \dfrac{\text{1.10 g}}{\text{1 mL}} = \text{110.0 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

         q₁      +           q₂         +       q₃      = 0

     nΔH      +       mCΔT      + C_calΔT = 0

0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0

0.0500ΔH +        2989       + 78.0       = 0

                             0.0500ΔH + 3067 = 0

                                          0.0500ΔH = -3067

                                                      ΔH = -3067/0.0500

                                                            = -61 000 J/mol

                                                            = -61 kJ/mol

\text{The enthalpy of reaction is $\large \boxed{\textbf{-61 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.

7 0
2 years ago
Give the ground-state electron configuration for each of the following elements. After each atom is its atomic number in parenth
Anni [7]

Answer:

(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

(b) ₁₀Ne: 1s² 2s² 2p⁶

---

(a) 3

(b) 6

(c) 7

Explanation:

We can state the ground-state electron configuration for each element following Aufbau's principle.

(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

(b) ₁₀Ne: 1s² 2s² 2p⁶

Second part

(a) Al belongs to Group 13 in the Periodic Table. It has 13-10=3 electrons in the valence shell.

(b) O belongs to Group 16 in the Periodic Table. It has 16-10=6 electrons in the valence shell.

(c) F belongs to Group 17 in the Periodic Table. It has 17-10=7 electrons in the valence shell.

3 0
3 years ago
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