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BartSMP [9]
4 years ago
11

What do all group 2 elements have in common?

Chemistry
2 answers:
Anestetic [448]4 years ago
8 0

Answer:

C is your answer

Explanation:

Nataly_w [17]4 years ago
3 0
They tend to form ionic bonds by losing electrons.
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1. Find your
bogdanovich [222]

Answer:

you can identify the properties of an element by studying its_____ on the ... between metals and nonmetals on the periodic table are the _____ which have ...

Explanation:

5 0
3 years ago
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white
Alex_Xolod [135]

The pH = 2.41

<h3>Further explanation</h3>

Given

5.0% by mass solution of acetic acid

the density of white  vinegar is 1.007 g/cm3

Required

pH

Solution

Molarity of solution :

\tt M=\dfrac{\%mass\times \rho\times 10}{MW~acetic~acid}\\\\M=\dfrac{5\times 1.007\times 10}{60}\\\\M=0.839

Ka for acetic acid = 1.8 x 10⁻⁵

[H⁺] for weak acid :

\tt [H^+]=\sqrt{Ka.M}

Input the value :

\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41

7 0
3 years ago
The atomic number is always equal to the atomic mass true false
kifflom [539]

Answer:

<h3>A neutral (no charge) particle found in the nucleus of an atom. ... The atomic number of an atom is equal to the number of protons or neutrons? false. The atomic number is always equal to the atomic mass.</h3>
7 0
3 years ago
Need help soon as possible
anyanavicka [17]

Answer:

The oeriodic table have 8 groups

3 0
2 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
3 years ago
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