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3 years ago

How do you find how many electrons are in a atom?

Chemistry

1 answer:

seropon [69]3 years ago

4 0

Answer:

To determine how many total electrons there are, add the amount of charge to the atomic number. In this case, there are fewer protons than electrons. For example, N 3- has a -3 charge which means it has 3 more electrons than a neutral nitrogen atom. Nitrogen’s atomic number is 7, therefore this ion has 10 electrons.

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Explanation:

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Why can a liquid take shape of the bottom of its container

Anna71 [15]

The particles in a liquid are not stuck in fixed positions, which is why liquids flow to take the shape of a container into which they are placed. Hope this helps :D

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3 years ago

Read 2 more answers

What is △n for the following equation in relating Kc to Kp?

Nimfa-mama [501]

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1

7 0

4 years ago

CHEM HELP PLEASE (: How many grams do 1.3 x 10^21 atoms of sodium weigh?

Yuliya22 [10]

1.3 x 10^21 atoms Na is 4.63230769231 mol Na

4.63230769231 mol Na * 23g/mol Na = 106.543076923g

8 0

3 years ago

Mn(OH)2(s) + MnO4(aq) → MnO42–(aq) (basic solution) When the equation is balanced with smallest whole number coefficients, what

PilotLPTM [1.2K]

Answer:

Hi, the given equation has some missing parts. Actual equation is- ' $Mn(OH)_{2}(s)+MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$ '

balanced equation: $Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Explanation:

$Mn(OH)_{2}(s)\rightarrow MnO_{4}^{2-}(aq.)$

Balance O and H in basic medium: $Mn(OH)_{2}(s)+6OH^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Balance charge: $Mn(OH)_{2}(s)+6OH^{-}(aq.)-4e^{-}\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$ ........(1)

$MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$

Balance charge: $MnO_{4}^{-}(aq.)+e^{-}\rightarrow MnO_{4}^{2-}(aq.)$ .....(2)

$[equation(2)\times 4]+[equation (1)]:$

$Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

$OH^{-}(aq.)$ is present on the left hand side of balanced equation and it's coefficient is 6

6 0

3 years ago

A 0.5895-g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid then needs 19

Zepler [3.9K]

Answer:

91.38 %

Explanation:

First we use the following formula to calculate the number of moles of HCl and NaOH contained in the solutions:

molar concentration = number of moles / solution volume

number of moles = molar concentration × solution volume

number of moles of HCl = 0.2050 × 100 = 20.5 mmoles

number of moles of NaOH = 0.1020 × 19.85 = 2.02 mmoles

reaction between NaOH and HCl:

NaOH + HCl → NaCl + H₂O

If 2.02 mmoles of NaOH were used for neutralization of HCl left from the reaction, which are 2.02 mmoles of HCl, it means that the HCl reacted with Mg(OH)₂:

20.5 - 2.02 = 18.48 mmoles of HCl

2 HCl + Mg(OH)₂ = MgCl₂ + 2 H₂O

From the chemical reaction we devise the following reasoning:

if 2 mmoles of HCl react with 1 mmole of Mg(OH)₂

then 18.48 mmoles HCl react with X mmoles of Mg(OH)₂

X = (18.48 × 1) / 2 = 9.24 mmoles of Mg(OH)₂

now the mass of Mg(OH)₂:

mass = number of moles × molecular weight

mass of Mg(OH)₂ = 9.24 × 58.3 = 538.7 mg = 0.5387 g

Now the percent by mass of magnesium hydroxide Mg(OH)₂ in the impure sample (purity):

purity = (mass of pure Mg(OH)₂ / mass of impure sample) × 100

purity = (0.5387 / 0.5895) × 100

purity = 91.38 %

7 0

4 years ago