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3 years ago

How are metals and nonmetals similar

Chemistry

1 answer:

Elza [17]3 years ago

6 0

Answer:

As compared to metals, they have low density and will melt at low temperatures. The shape of nonmetals cannot be changed easily because they are brittle and will break. Elements that have properties of both metals and nonmetals are called metalloids. They can be shiny or dull and their shape is easily changed.

Explanation:

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Use the information below to explain why the atomic radius decreases down a group.

notsponge [240]

Answer:

Detail is given below

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm

In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N

In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.

3 0

3 years ago

Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you ne

katrin [286]

Answer:

7.08

Explanation:

To solve this problem we'll use the Henderson-Hasselbach equation:

pH = pka + log $\frac{[A^-]}{[HA]}$

Where $\frac{[A^-]}{[HA]}$ is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

pka = -log (1.8x10⁻⁴) = 3.74

We input the data:

4.59 = 3.74 + log $\frac{[A^-]}{[HA]}$

And solve for $\frac{[A^-]}{[HA]}$ :

0.85 = log $\frac{[A^-]}{[HA]}$

$10^{(0.85)}$ = $\frac{[A^-]}{[HA]}$

$\frac{[A^-]}{[HA]}$ = 7.08

3 0

3 years ago

Three blocks of the same mass are placed in front of you. Block A has a volume of

Ivan

Given :

Three block of same mass name A , B and C .

Sides of block A , B and C is 3.0 cm , 5.0 cm and 10.0 cm .

To Find :

Which block has the higher density .

Solution :

We know , density $\rho$ is given by :

$\rho=\dfrac{M}{V}$ ......1 )

Here , V is volume .

Now , volume V for cube is given :

$V=a^3$ .......2 )

( Here , a is the side of cube )

Now ,form equation 1 we can see if mass remains constant then density decrease with increase in volume.

Therefore , cube with minimum side will have higher density , in this case it is 3 cm block .

Hence , this is the required solution .

8 0

3 years ago

Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the

KengaRu [80]

Answer : The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

$Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}$

The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

5 0

3 years ago

A 100 g sample of an unknown liquid absorbs 2000 j of heat energy, raising the liquid's temperature from 50 ◦ c to 70 ◦

Mekhanik [1.2K]

Since there is no phase change, we can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J kg⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Q = 2000 J
m = 100 g = 0.1 kg
c = ?
ΔT = (70 °C - 50 °C) = 20 °C

By applying the formula,
2000 J = 0.1 kg x c x 20 °C
c = 2000 J / (0.1 kg x 20 °C)
c = 1000 J kg⁻¹ °C⁻¹

Hence, the specific heat capacity of the liquid is 1000 J kg⁻¹ °C⁻¹.

5 0

4 years ago