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love history [14]
3 years ago
14

( urgent i have limited time ! please help me solve i'll give brainlist )

Chemistry
1 answer:
Xelga [282]3 years ago
7 0

Answer:

where are those two images which you have sent

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18 g of copper is mixed with silver nitrate in water. How much copper ll nitrate will formed?
Lorico [155]

Answer:

Mass = 112.54 g

Explanation:

Given data:

Mass of copper = 18 g

How much copper(II) nitrate formed = ?

Solution:

Cu + 2AgNO₃  →  Cu(NO₃)₂ + 2Ag

Number of moles of copper:

Number of moles = mass/ molar mass

Number of moles = 18 g/ 29 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of Cu with Cu(NO₃)₂ .

             Cu        :        Cu(NO₃)₂

              1           :             1

           0.6          :           0.6

Mass of Cu(NO₃)₂ :

Mass = number of moles × molar mass

Mass = 0.6 mol × 187.56 g/mol

Mass = 112.54 g

7 0
3 years ago
Ba(oh)2+H3po4+h2o how is it <br> balance ?
Igoryamba

Answer:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O

Explanation:

Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O

There are 3 atoms of Ba on the right side and 1atom on the left side. It can be balance by putting 3 in front of Ba(OH)2 as shown below:

3Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O

There are 2 atoms of P on the right side and 1atom on the left. It can be balance by putting 2 in front of H3PO4 as shown below:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + H2O

Now, there are a total of 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O

Now the equation is balanced as the numbers of the atoms of the different elements present on both sides are equal

4 0
3 years ago
Which expression represents a reaction rate?a. time/massc. energy/timeb. number/timed. time/energy
Kitty [74]
The right answer for the question that is being asked and shown above is that: "<span>b. number/timed." Reaction Rate refers to the </span> speed of reaction<span> for a reactant or product in a particular </span>reaction<span> is intuitively defined as how fast or slow a</span>re action<span> takes place.</span>
4 0
3 years ago
Read 2 more answers
Which of the following best describes the forces that generate solar power
Sergeeva-Olga [200]
I believe the answer is A (it was on another Brainly question as well).
8 0
3 years ago
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
3 years ago
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