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3 years ago

13

The empirical formula of glucose is CH2O, and the molecular formula of glucose is 6 times than the empirical formula. What is th

e molecular formula of glucose?

1 answer:

goldfiish [28.3K]3 years ago

3 0

C6H12O6
That is the molecular formula of glucose

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Identify the products and the reactants in the following chemical equations

liq [111]

Answer:

1-Mg+O2=reactants MgO-product

2-CaNr2+AlCl3=reactants CaCI2+AIBr3=product

3-C2H4+O2=reactants H2O+CO2=product

Explanation:

Reactants are on the left side and products on the right.

6 0

1 year ago

Who is Gregor Mendeſ, and what did he<br> discover about traits?

shtirl [24]

Answer:

Gregor Mendel, through his work on pea plants, discovered the fundamental laws of inheritance. He deduced that genes come in pairs and are inherited as distinct units, one from each parent. Mendel tracked the segregation of parental genes and their appearance in the offspring as dominant or recessive traits.

Through his careful breeding of garden peas, Gregor Mendel discovered the basic principles of heredity and laid the mathematical foundation of the science of genetics.

Explanation:

8 0

2 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

2 years ago

An ion with 11 protons 12 neutrons and a charge of + 1 has an atomic number of

posledela

Answer:

11 electrons total and 23amu

Explanation:

This tells us that sodium has 11 protons and because it is neutral it has 11 electrons. The mass number of an element tells us the number of protons AND neutrons in an atom (the two particles that have a measurable mass). Sodium has a mass number of 23amu.

5 0

2 years ago

A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

$k=9.98\times 10^{-2}hr^{-1}$

Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

8 0

2 years ago