All categories

2 years ago

What allowed Mendeleev to make predictions of undiscovered elements

2 answers:

7 0

Predictions using gaps Mendeleev left gaps in his table to place elements not known at the time. By looking at the chemical properties and physical properties of the elements next to a gap, he could also predict the properties of these undiscovered elements.

Andrej [43]2 years ago

3 0

Answer:

he predicted the properties from known elements above and belws the unknown in the same group

Explanation:

What allowed Mendeleev to make predictions of undiscovered elements

He realized that an element on this table with one known element above it and one known element below it had to have properties between the two known elements

How did Mendeleev predict gallium and germanium?

Based on gaps in the periodic table Mendeleev deduced that in these gaps belonged elements yet to be discovered. Based on other elements below and above in the same group he predicted the existence of eka-aluminum, eka-boron, and eka-silicon, later to be named gallium (Ga), scandium (Sc), and germanium (Ge).

You might be interested in

when riding bumper cars at the fair you bump into your friend and your car bounces backwards which law of motion is this

weeeeeb [17]

Newton's third law of interaction, says that if one body exerts a force on a second body, the second body exerts a force equal in magnitude and opposite in direction on the first body. It's the law of action-reaction, and it helps to explain why you feel a jolt when you collide with another bumper car.

6 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

What are the coefficients for the reaction _Cl2O5 + _H2O → _HClO3 once it is balanced?

Eva8 [605]

It is B (1,1,2) because when you plug those numbers in, there are the same amount of elements in the reactant and product, just rearranged as the format A+B → AB, because this is a synthesis reaction.

7 0

2 years ago

Read 2 more answers

Which statement most accurately compares the temperature of the core of the sun to its surface?

DochEvi [55]

C) the core is more than a million times hotter than the surface

3 0

2 years ago

The heat required to raise the temperature of 12g of water from 16 C to 21 C is:

NemiM [27]

Answer:

The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

Q=?
c= 4.186 $\frac{J}{g*C}$
m= 12 g
ΔT=Tfinal - Tinitial= 21 °C - 16°C= 5 °C

Replacing:

Q= 4.186 $\frac{J}{g*C}$ *12 g *5 °C

Solving:

Q=251.16 J

Since 1 J is equal to 0.2388 cal, then the following rule of three can be applied: if 1 J is equal to 0.2388 cal, then 251.16 J to how many cal are?

$cal=\frac{251.16 J * 0.2388 cal}{1 J}$

cal= 59.98 ≅ 60

The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.

3 0

2 years ago