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AnnyKZ [126]
3 years ago
15

Which of these statements best describes an Alkali? *

Chemistry
1 answer:
lana66690 [7]3 years ago
5 0
Proton donor. Alkali metals have one electron that they want to give up to fulfill the octet rule (makes them stable)
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2. What is the chemical name for GaC1z?
VMariaS [17]

Answer:

aqueous gallium chloride i think

Explanation:

3 0
3 years ago
A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
Anika [276]
The heat that is required to raise the temperature of an object is calculated through the equation,
                        heat = mass x specific heat x (T2 - T1)
Specific heat is therefore calculated through the equation below,
                                specific heat = heat / (mass x (T2 - T1))
Substituting,
                                specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The value of specific heat from above equation is 0.838 J/g°C. 
5 0
3 years ago
An atom of element A has 108 protons. What would the elements atomic number be?
marusya05 [52]

108 is the atomic number

3 0
3 years ago
What is the volume, in liters, of 576 grams of so2 gas at stp?
katen-ka-za [31]
The volume  in  liters  of  576 grams  of SO2  gas  at STP  is   calculated as below

calculate the moles of SO2 = mass/molar  mass

= 576 g/64 g /mol  = 9 moles

At  STP  1mole =22.4 L
what  9 mole =? liters

by cross  multiplication

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5 0
3 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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