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-BARSIC- [3]
3 years ago
8

The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this

information?
a. They are in the same group because they have similar atomic masses, but they are in different periods because they have different chemical properties.

b. They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

c. They are in the same group because their atomic numbers are each multiples of four, but they are in different periods because they have different chemical properties.

d. They are in the same group because they have similar atomic masses, but they are in different periods because they have very different atomic numbers.
Chemistry
2 answers:
Cloud [144]3 years ago
8 0
The answer is b) similar chemical properties but very different atomic numbers
SSSSS [86.1K]3 years ago
4 0

Answer:

B: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

Explanation:

I took the Unit test on EDG and I got it correct

Pls mark me brainliest

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salantis [7]

Answer:

1. Uniform, non visible

2. Visible, non uniform

3. Uniform, non visible

4. Visible, non uniform

5. Visible, non uniform

Explanation:

A heterogeneous mixture is simply any mixture that is not uniform in composition — it's a non-uniform mixture of smaller constituent parts. By contrast, a mixture that is uniform in composition is a homogenous mixture

visible state of matter is when you can easily difference two components of a mixture

non visible is when the state of the matter in the substance is not visible to the eye, it can't be differentiated

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How many grams are in CO2?
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There are 44grams in CO2
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What is the metric unit for mass?<br> a.Meters<br> b.Kilograms<br> c.Pounds<br> d.matter
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The metric unit for mass is B. Kilograms
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4 years ago
What is concentration​
Arlecino [84]

Answer:

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6 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
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