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1 year ago

11

Use the periodic table to select which type of bond is present and which of the listed properties is most likely for each substa

nce.
A =
B =
Cu3Zn2
O2

1 answer:

hoa [83]1 year ago

4 0

Answer:

A :metallic

B :conducts electricity as a solid

C :covalent

D :does not conduct electricity

Explanation:

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Beaker A contains 2.06 mol of copper ,and Barker B contains 222 grams of silver.Which beaker the larger number of atom?

Dmitry [639]

Answer:

The number of copper atoms 12.405 ×10²³ atoms.

The number of silver atoms 13.13 ×10²³ atoms.

Beaker B have large number of atoms.

Explanation:

Given data:

In beaker A

Number of moles of copper = 2.06 mol

Number of atoms of copper = ?

In beaker B

Mass of silver = 222 g

Number of atoms of silver = ?

Solution:

For beaker A.

we will solve this problem by using Avogadro number.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.

While we have to find the copper atoms in 2.06 moles.

So,

63.546 g = 1 mole = 6.022×10²³ atoms

For 2.06 moles.

2.06 × 6.022×10²³ atoms

The number of copper atoms 12.405 ×10²³ atoms.

For beaker B:

107.87 g = 1 mole = 6.022×10²³ atoms

For 222 g

222 g / 101.87 g/mol = 2.18 moles

2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms

8 0

2 years ago

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

<u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

1 year ago

A 5-ounce cup of raspberry yogurt contains 5.0 g of protein, 3.0 g of fat, and 13.3 g of carbohydrate. The fuel values for prote

castortr0y [4]

Answer:

as it has 17000 joules and 1 calorie has 4184 joules, this yogurt serving has 4.06 calories

6 0

1 year ago

A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq)

AnnyKZ [126]

Answer:

$\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}$

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

0.0150 x

$K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}$

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

x 2.293 × 10⁻⁵

$K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}$

4 0

1 year ago

A tank of oxygen has a volume of 40.0L and is held at a pressure of 159atm at 25∘C. What volume of O2 gas (in liters) would ther

Advocard [28]

Answer:

V2 = 6616 L

Explanation:

From the question;

Initial volume = 40L

Initial Pressure, P1 = 159atm

Initial Temperature T1 = 25 + 273 = 298K (Upon converting to Kelvin unit)

Final Volume, V2 = ?

Final Pressure, P2 = 1 atm

Final Temperature T2 = 37 + 273= 310K (Upon converting to Kelvin unit)

These quantities are related by the equation;

P1V1 / T1 = P2V2 / T2

V2 = T2 * P1 * V1 / T1 * P2

V2 = 310 * 159 * 40 / (298 * 1)

V2 = 6616 L

6 0

1 year ago