Which of the samples would be classified as a mixture

Part A:

uranmaximum [27]

Answer:

Part A = The mass of sulfur is 6.228 grams

Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams

Explanation:

Part A

Step 1: Data given

A mixture of carbon and sulfur has a mass of 9.0 g

Mass of the product = 27.1 grams

X = mass carbon

Y = mass sulfur

x + y = 9.0 grams

x = 9.0 - y

x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6

(9 - y)*(44.01/12.01) + y(64.07/32.07)

(9-y)(3.664) + y(1.998)

32.976 - 3.664y + 1.998y = 22.6

-1.666y = -10.376

y = 6.228 = mass sulfur

x = 9.0 - 6.228 = 2.772 grams = mass C

The mass of sulfur is 6.228 grams

Part B

Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).

Calculate moles of 1 silver atom

Moles = 1/ 6.022*10^23

Moles = 1.66*10^-24 moles

Mass = moles * molar mass

Mass = 1.66*10 ^-24 moles *107.87

Mass = 1.79 * 10^-22 grams

The mass of 1 silver atom is 1.79 * 10^-22 grams

5 0

2 years ago

Read 2 more answers

Hi May I know how to balance this

almond37 [142]

Answer:

2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

Explanation:

Equating coefficients, you get ...

aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃

For Ba: 3a = d

For P: 2a = 4c

For O: 8a +2b = 10c +3d

For Si: b = d

__

Expressing everything in terms of b and c, we get ...

d = b

a = b/3 = 2c

From the second, b = 6c, so we have ...

a = 2c

b = 6c

c = c

d = 6c

And we can write the equation with c=1 as ...

2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

4 0

3 years ago

Draw the Lewis dot structure for RbIO2. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atom

Rzqust [24]

Answer : The Lewis-dot structure of $RbIO_2$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $RbIO_2$

As we know that rubidium has '1' valence electrons, iodine has '7' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $RbIO_2$ = 1 + 7 + 2(6) = 20

As we know that $RbIO_2$ is an ionic compound because it is formed by the transfer of electron takes place from metal to non-metal element.

7 0

3 years ago

Which of these best describes the bonding between oxygen and hydrogen in the water molecules?

FromTheMoon [43]

When oxygen and hydrogen combine they form a polar covalent bond

7 0

3 years ago

Read 2 more answers

On the basis of the information above, a buffer with a pH = 9 can best be made by using

telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for $\mathbf{H_3PO_4 , H_2PO^{-}_4 , HPO_4^{2-}}$ are $\mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \ 5\times 10^{-13}}$ respectively.

$\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}$

$\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}$

$\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}$

The reason while option D is the best answer is that, the value of pKa for both

$\mathbf{H_2PO^{-}_4 ,\ \& \ HPO_4^{2-}}$ lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

$\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}$

3 0

2 years ago