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3 years ago

How many protons and electrons are present in the mg2+ ion?

1 answer:

7 0

There are 12<span> protons and </span>10<span> electrons in a </span><span>Mg<span>2+</span></span><span> ion, the normal amount of neutrons is </span>12<span>.</span>

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What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks

Firlakuza [10]

Answer:

$\large \boxed{1.64\times 10^{-5}\text{ mol/L }}$

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

2x 0.007 50 + x

$K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}$

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3 years ago

The more the energy, the larger the (a,b,c,d)?

Elanso [62]

Answer: D:wavelenght

Explanation: Students will understand that shorter wavelengths have higher frequency and energy.

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3 years ago

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All of the following are examples of minerals except __________.

snow_tiger [21]

The answer is D.timber good luck

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3 years ago

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A 0.0200 gram piece of unknown alkaline earth metal, M, is reacted with excess 0.500 M H 2 SO 4 , and the hydrogen gas produced

Rama09 [41]

Answer:

0.981atm

Explanation:

According ot Dalton's law total pressure of a mixture of non-reactive gas is equal to sum of partial pressures of individual gases.

total pressure= 1.01at

Number of gases=2

Gases: water vapor and hydrogen

partial pressure of water vapor= 0.029atm

1.01= partial pressure of water vapor+ partial pressure of hydrogen

1.01= 0.029 + partial pressure of hydrogen

partial pressure of hydrogen = 0.981atm

6 0

3 years ago

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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1 year ago