What type of map should have a north arrow, key ,and map scales?

PLEASEEEEE HELP MATH EXPERTS<br><br> What is the value of x?

diamong [38]

X=107 because the sides of a hexagon must add up to 720

6 0

2 years ago

A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?

zhannawk [14.2K]

POH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁹ ]

pOH = 9

Answer C

hope this helps!

3 0

2 years ago

What kind of intermolecular forces act between a sodium cation and an ammonia molecule? Note: If there is more than one type of

Reil [10]

Answer:

Ion-dipole forces

Explanation:

Na⁺ is a cation, that is, an ion with a positive charge.

NH₃ has polar covalent bonds (due to the difference in electronegativity between nitrogen and hydrogen). According to the VESPR theory, it has a trigonal pyramidal shape with a lone pair. As a consequence, it has a net dipole moment and the molecule is polar.

The intermolecular forces between Na⁺ (ion) and NH₃ (dipole) are ion-dipole forces.

7 0

2 years ago

Which statement describes heterogeneous mixtures?

Artemon [7]

<span>The righ answer is the option A. They are made up of two or more pure substances that are not chemically bonded. A classicall expample ot heterogeneous mixtures are sugar and salt. Sugar is a pure substance, salt is also a pure substance, when you mixe them you form mixture where salt is still salt and sugar is still sugar, that is what "they are not chemically bonded" means. So you can separate them by physical media. </span>

4 0

2 years ago

Read 2 more answers

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

2 years ago