# What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL

The volume of 45.6g of silver if the density of silver is 10.5g/mL is 4.342 ml. The correct option is A.

### What is Volume?

Volume is the space occupied by a three-dimensional object. The volume of any object can be calculated by dividing the mass by its density. It is a scalar quantity. It is the total weight is that object.

Silver is an element in the periodic table. It is non-metal, and it is used in making ornaments and in medicines. The volume of the solver is calculated, and the mass and density are given.

The mass of silver is given, 45.6g

The density of the element is 10.5g/mL

Putting the value in the equation

The density and the mass would be divided.

Volume = 45.6g / 10.5 = 4.342

Thus, the volume of silver is 4.342 ml. The correct option is A.

brainly.com/question/2472349

#SPJ2

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

## Related Questions

How many moles of argon are in 4.80x10^24 atoms of argon

7.97 mol Ar

Explanation:

Use Avogadro's number to convert atoms to moles.  This number is the number of atoms in one mole.  There are 6.022 × 10²³ atoms in one mole.

Divide the number of atoms given by Avogadro's number.

(4.80 × 10²⁴)/(6.022 × 10²³) = 7.97 mol

There are 7.97 moles of argon.

3.0 moles  (A P E X)

Explanation:

Which one of these could be in the unknown anion "X" in this acid: H3Xcarbonate
fluorate
nitrogen
nitrite

Could you explain how to find this? The process?

the answer is nitrogen

What is colloidal solutions

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid.

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. If 100 g of nitrogen react with 6 g of hydrogen, how much ammonia will be produced? 106 g 34 g 128 g 40 g

34 g

Explanation:

Let's consider the following balanced equation.

N₂ + 3 H₂ → 2 NH₃

The theoretical mass ratio of N₂ to H₂ is 28g N₂ : 6g H₂ = 4.6g N₂ : 1g H₂.

The experimental mass ratio of N₂ to H₂ is 100g N₂ : 6g H₂ = 16.6g N₂ : 1g H₂.

As we can see, hydrogen is the limiting reactant.

According to the task, we 6 g of H₂ react completely, 34 g of ammonia are produced.

The SI unit of time is the second, which is defined as 9,192,631,770 cycles of radiation associated with a certain emission process in the cesium atom. Calculate the wavelength of this radiation (to three significant figures).

Given:

Radiation emission in Cs atom = 9,192,632,770 cycles

To determine:

The wavelength of the above radiation

Explanation:

It is given that :-

1 sec equivalent to 9,192, 631, 770

Now, frequency (ν) = cycles /sec = 9,192, 631, 770/sec

Wavelength of a radiation is given as:

λ = c/ν

where c = speed of light = 3*10⁸ m/s

λ = 3*10⁸ ms⁻¹/9,192, 631, 770 s⁻¹ = 0.0326 m

Ans: Thus the wavelength of this radiation is 0.033 m

To calculate the wavelength of this radiation if the SI unit of time is the second, which is defined as 9,192,631,770 cycles of radiation associated with a certain emission process in the cesium atom, the wavelength is one cycle of radiation, and therefore the wavelength is 1/9192631770

The electronic structure 1s22s22p63523p64523d8 refers to the ground state ofKr.
Ni.
Pd.
Fe.
none of the above.
Previous

The given electronic configuration 1s²2s²2p⁶3s²3p⁶4s²3d⁸ is the Nickel (Ni). Therefore, option (B) is correct.

### What is an electronic configuration?

The electron configuration can be explained as electrons being filled in different energy levels of an atom of a particular element. In the electron configuration, the number of electrons is usually written as a superscript of atomic subshells. For example, the electron configuration of Oxygen is written as 1s²2s²2p⁴.

The sequence of fully filled subshells that correspond to the electronic configuration of a noble gas is denoted by square brackets. The principal quantum number (n) can be used to evaluate the maximum number of electrons in an electron shell.

The value of 2n² is given the maximum number of electrons. The atomicorbitals at low energy must be filled first occupying an orbital with a higher energy level.

The total number of electrons filled in the given electronicconfiguration  1s²2s²2p⁶3s²3p⁶4s²3d⁸  is 28. The atomic number of Nickel is 28 therefore it is the configuration of Nickel.