ALGEBRA 1 SEM 1

How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl2 according to the balan

Mashutka [201]

Answer:

6.55 mL of K₃PO₄ are required

Explanation:

We need to propose the reaction, in order to begin:

2K₃PO₄ (aq) + 3NiCl₂(aq) → Ni₃(PO₄)₂ (s) ↓ + 6KCl (aq)

Molarity = mol/L (Moles of solute that are contained in 1 L of solution.)

M = mol / volume(L). Let's find out the moles of chloride:

- We first convert the volume from mL to L → 187 mL . 1L / 1000mL = 0.187L

0.0184 M . 0.187L = 0.00344 moles of NiCl₂

Ratio is 3:2. Let's propose this rule of three:

3 moles of chloride react with 2 moles of phosphate

Then, 0.00344 moles of NiCl₂ will react with (0.00344 . 2) /3 = 0.00229 moles of K₃PO₄

M = mol / volume(L) → Volume(L) = mol/M

Volume(L) = 0.00229 mol / 0.350 M = 6.55×10⁻³L

We convert the volume from L to mL → 6.55×10⁻³L . 1000mL /1L = 6.55 mL

5 0

3 years ago

How many grams of carbon dioxide would be formed if 38.9 g C2H2 reacted completely with oxygen

raketka [301]

2C₂H₂ + 5O₂ = 4CO₂ + 2H₂O

m(CO₂)/{4M(CO₂)} = m(C₂H₂)/{2M(C₂H₂)}

m(CO₂)=2M(CO₂)m(C₂H₂)/M(C₂H₂)

m(CO₂)=2*44g/mol*38.9g/26g/mol = 131.7 g

131.7 grams of carbon dioxide would be formed

6 0

3 years ago

Read 2 more answers

A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a l

elixir [45]

Answer:

$T_{C}$ = -4.2°C

$T_{H}$ = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

| $\frac{Q_{C} }{Q_{H} }$ | = $\frac{T_{C} }{T_{H} }$

Similarly,

| $Q_{H}$ | = | $Q_{C}$ | + | $W_{S}$ |

During winter, the value of | $T_{H}$ | = 20°C = 273.15 + 20 = 293.15 K and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{H}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{H} }$ | = 1 - $\frac{T_{C} }{T_{H} }$

1.5/0.75*(293.15- $T_{C}$ ) = 1 - ( $T_{C}$ /293.15

Further simplification,

$T_{C}$ = -4.2°C

During summer, $T_{C}$ = 25°C = 273.15+25 = 298.15 K, and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{C}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{C} }$ | = $\frac{T_{H} }{T_{C} }$ - 1

1.5/0.75*( $T_{H}$ - 298.15) = ( $T_{H}$ /298.15

Further simplification,

$T_{H}$ = 49.4°C

4 0

4 years ago

Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a

Ghella [55]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule is gotten according to longest carbon chain = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

3 0

3 years ago

What is the periodic trend in atomic size and electronegativity as you move from left to right on the?

Darya [45]

As you move from left to right on the periodic table, atomic size and electronegativity increase.

(Ionization energy increases from left to right on the periodic table, and electronegativity relates to ionization energy in that the higher the ionization energy, the higher the electronegativity.)

(Atomic radius increases from left to right on the periodic table due to the fact that electrons are gradually being added from left to right, meaning the amount of protons in the nucleus are increasing in order to keep the atom neutral, therefore atomic size/radius increases.)

6 0

3 years ago