Define a table salt in salty water A). Solution B). Solute C). Solvent D). Salute

Draw and label the parts of a helium atom. Include the mass and charge of each subatomic particle.

mihalych1998 [28]

Explanation:

Helium is the second element on the periodic table.

Information about helium:

it belongs to group O on the periodic table
it is has an atomic number of 2
Helium exists naturally as gas and it is nonreactive.

To write an atom we use this format:

ₙᵇGˣ

G is the symbol of the atom

n is the atomic number

b is the mass number

x is the charge on the atom

Using the periodic table as guide

Symbol of helium is He

Atomic number of helium is 2

mass number of helium is 4

charge on helium atom is 0

⁴₂He

Every atom contains protons, neutrons and electrons;

Electrons are negatively charged. For a neutral atom, the number of electrons is the same as that of protons.

Protons are positively charge particles in an atom. The atomic number is the number of protons in an atom.

neutrons do not have charges.

Helium has:

Number of protons = 2

Number of electrons = 2

Number of neutrons = 2

Mass of each subatomic particle:

1 electron = 9.11 x 10⁻³¹kg

2 electrons = 1.82 x 10⁻³⁰kg

Protons and neutrons have the same mass:

1 proton = 1.67 x 10⁻²⁷kg

2 protons = 3.34 x 10⁻²⁷kg

2 neutrons = 3.34 x 10⁻²⁷kg

Learn more:

helium brainly.com/question/2439349

#learnwithBrainly

4 0

3 years ago

What is the molarity of the solution resulting from the dissolution of 239 g glucose (C6H12O6) in 250

kifflom [539]

Answer:

Molarity =5.32 M

Explanation:

Given data:

Mass of glucose = 239 g

Volume = 250 mL (250 /1000 = 0.25 L)

Molarity = ?

Solution;

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 239 g / 180.2 g/mol

Number of moles = 1.33 mol

Molarity:

Molarity = number of moles / volume in litter

Molarity = 1.33 mol / 0.25 L

Molarity =5.32 M

6 0

3 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

2 years ago

In the addition of hbr to 1-butyne the electrophile in the first step of the mechanism is

irga5000 [103]

Answer:
In the addition of hbr to 1-butyne the electrophile in the first step of the mechanism is Hydrogen atom of HBr.

Explanation:
In this reaction first of all HBr approaches the triple bond. A Pi Complex (weak inter-molecular interactions) is formed between the two molecules. And the triple bond attacks the partial positive hydrogen atom creating a negative charge on Bromine along with positive charge on itself (Sigma Complex). In second step the negative Bromide attacks the positive carbon of Butyne.

7 0

2 years ago

The ka of hypochlorous acid (hclo) is 3.0 ⋅ 10−8 at 25.0 °c. calculate the ph of a 0.0375m hypochlorous acid solution.

Scrat [10]

We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375    0 0
Change    -x     +x +x
Equilibrium 0.0375-x    x x

We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)

Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47

6 0

3 years ago

Define a table salt in salty waterA). SolutionB). SoluteC). SolventD). Salute​

Define a table salt in salty waterA). SolutionB). SoluteC). SolventD). Salute