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3 years ago

An atom of a mystery element contains 6 protons, 6 electrons, and 8 neutrons.

Chemistry

1 answer:

Leno4ka [110]3 years ago

8 0

Answer:

Explanation:

Mass number is just the value of the protons and neutrons added, in this case its 14

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padilas [110]

Lol now way he answers this

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3 years ago

summer school is over, and I'm feeling self-destructive. Take all of my points and have a good night. 3/3

Vanyuwa [196]

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3 years ago

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Can an Atom be broken down into a smaller substance? Yes or No Explain.

Pepsi [2]

Yes it can as it’s made of protons neutrons and electrons which is how we break the atoms down

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3 years ago

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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

3 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

4 years ago