CHEMISTRY HIGH SCHOOL

Atom X has a total of 52 elementary particles and a mass number of 35. X's atomic number is

Answers

Answer 1

Answer:

Answer:

Explanation:

The number of elementary particles in the atom X = 52

The mass number of atom X = 35

The information required = The atomic number

The number elementary particles of the atom = The number of protons + The number of neutrons + The number of electrons

The mass number of an atom = The number of protons + The number of neutrons

∴ The number of electrons = The number elementary particles of the atom - The number elementary particles of the atom

The number of electrons = 52 - 35 = 17

In a neutral atom, the number of protons = The number of electrons = 17

The atomic number of atom, X = The number of protons in X = 17

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Molarity. 0.5 GRAMS of sodium chloride is dissolved to make 0.05 liters of solution.

would it be the same as..

0.5 MOLES of sodium chloride is dissolved to make 0.05 liters of solution?

please help me with the grams. thank you (: first time using this!

Answers

Find if the same solution would be equal:
=> 0.5 grams of Sodium chloride dissolved in 0.05 liters solution
=> 0.5 moles of sodium chloride dissolved in 0.05 liters solution
The answer is NO. They are not equal because 0.5 grams of sodium chloride is equivalent to 10 moles of sodium chloride which makes 0.5 grams of sodium chloride mixed with 0.05 liters of solution more concentrated than 0.5 moles of sodium chloride dissolved in 0.05 liters of solution.

An environmental chemist needs a carbonate buffer of pH 10.00. How many grams of Na,Co, must she add to 1.5 L of 0.20 M NaHCO,? K, of HCO, = 4.7 x 10"

Answers

Answer: a

Explanation: it just like simple math but it just adding stuff

Draw .. the structural formula ( using c and h ) for 2- pentyne

Answers

Answer : The structural formula of 2-pentyne is shown below.

Explanation :

Structural formula : In the structural formula, the bonding and type of bonds which holds the atoms in molecule together are shown.

The given name of compound is, 2-pentyne.

A suffix '-yne' present at the end of the name represent the alkyne group.

For the number of carbon atom, we use prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.

In this compound, the parent chain is 5 membered and a triple bond is present at 2nd position of the parent chain.

Thus, the structural formula of 2-pentyne is shown below.

answer is below.........

How did Moseley establish a more accurate periodic table?

Answers

Moseley established a more accurate periodic table by closer determination of atomic numbers. Through the results of his measurements of the wavelengths of the X-ray spectral lines of a number of elements which showed that the ordering of the elements by atomic number.

Answer:

B. by arranging the elements according to atomic number instead of atomic mass

Explanation:

i just took the test

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water (d = 1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50 % v/v (b) mass percent 52.7 % w/w (c) molarity M (d) molality m (e) mole fraction

Answers

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

$\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}* \text{Volume of ethylene glycol}=1.114g/mL* 1mL=1.114g$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1.00g/mL* 1mL=1.00g$

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(2.114g)/(1.070g/mL)=1.975mL$

(a) Now we have to calculate the volume percent.

$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}* 100=(1mL)/(1.975mL)* 100=50.63\%$

(b) Now we have to calculate the mass percent.

$\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}* 100=(1.114g)/(2.114g)* 100=52.69\%$

(c) Now we have to calculate the molarity.

$\text{Molarity}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Volume of solution (in mL)}}$

$\text{Molarity}=(1.114g* 1000)/(62.07g/mole* 1.975L)=9.087mole/L$

(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Mass of water (in g)}}$

$\text{Molality}=(1.114g* 1000)/(62.07g/mole* 1kg)=17.947mole/kg$

(e) Now we have to calculate the mole fraction of ethylene glycol.

$\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}$

$\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=(1.114g)/(62.07g/mole)=0.01795mole$

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=(1g)/(18g/mole)=0.0555mole$

$\text{Mole fraction of ethylene glycol}=(0.01795mole)/(0.01795mole+0.0555mole)=0.244$

What sequence of enzymes allows for the synthesis of glycogen from glucose-6-phosphate?a. UDP-glucose pyrophosphorylase, phosphoglucomutase, glycogen synthase, amylo-(1,4α1,6)-transglycosylase

b. phosphoglucomutase, UDP-glucose pyrophosphorylase, amylo-(1,4α1,6)-transglycosylase, glycogen phosphorylase

c. phosphoglucomutase, amylo-(1,4α1,6)-transglycosylase, UDP-glucose pyrophosphorylase, glycogen synthase

d. UDP-glucose pyrophosphorylase, phosphoglucomutase, amylo-(1,4α1,6)-transglycosylase, glycogen synthase

e. phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase, amylo-(1,4α1,6)-transglycosylase

Answers

Answer:

The answer is (e) : phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase then amylo-(1,4-1,6)-transglycosylase.

Explanation:

Phosphoglucomutase: Convert glucose-6-phosphate to glucose-1-phosphate.

UDP-glucose pyrophosphorylase: Form UDP-glucose from glucose-1-phosphate.

Glycogen synthase: Add the new glucose from UDP-glucose to the growing glycogen chain.

Amylo-(1,4-1,6)-transglycosylase: This is a branching enzyme, it initiates formation of branches evolving from the main chain.

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