2. Calculate the mass of 5.35 mole of H2O2.

Dissolving a spoonful of sugar in tea or coffee is an example of

Virty [35]

It’s an example of dissolving

5 0

2 years ago

How many significant figures are in this number? 107.051

Anestetic [448]

Answer:

there are 6 significant figures in 107.051

7 0

3 years ago

Read 2 more answers

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io

Dmitrij [34]

Answer :

(a) The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

The spectator ions are, $NH_4^{+}\text{ and }SO_4^{2-}$

(b) The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

(c) The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

$Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)$

The ionic equation in separated aqueous solution will be,

$2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)$

In this equation, $NH_4^{+}\text{ and }SO_4^{2-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

(b) The given balanced ionic equation will be,

$Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)$

The ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

(c) The given balanced ionic equation will be,

$Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)$

The ionic equation in separated aqueous solution will be,

$Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

3 0

3 years ago

How does the number of valence electrons determine the type and number of bonds an element forms?

Advocard [28]

The number of electrons in an atom's outermost valence shell governs its bonding behaviour. Elements whose atoms have the same number of valence electrons are grouped together in the Periodic Table. ... Nonmetals tend to attract additional valence electrons to form either ionic or covalent bonds.

8 0

3 years ago

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

The volume of the gas is 2.80 L.

7 0

3 years ago