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ss7ja [257]
3 years ago
5

A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou

les what is the approximate final temperature of the water.
74 C
78C
81 C
83C
Chemistry
1 answer:
Vinvika [58]3 years ago
3 0

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

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gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

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3 years ago
Define the following symbols that are encountered in rate equations: [A]0, t1/2 [A]t, k.
dezoksy [38]
[A]0= Initial concentration
t1/2= half life
[A]= final concentration
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5 0
3 years ago
Consider the reaction CaCN2 + 3 H2O → CaCO3 + 2 NH3 . This reaction has a 75.6% yield. How many moles of CaCN2 are needed to obt
kherson [118]

Answer: Thus 0.724 mol of CaCN_2 are needed to obtain 18.6 g of NH_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} NH_3=\frac{18.6g}{17g/mol}=1.09moles

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

According to stoichiometry :

2 moles of NH_3 are produced by = 1 mole of CaCN_2

Thus 1.09 moles of NH_3 will be produced by =\frac{1}{2}\times 1.09=0.545moles  of CaCN_2

But as yield of reaction is 75.6 %, the amount of CaCN_2 needed is =\frac{0.545}{75.6}\times 100=0.724

Thus 0.724 mol of CaCN_2 are needed to obtain 18.6 g of NH_3

3 0
3 years ago
How much time does it take for light from the center of the galaxy to reach Earth?
Ede4ka [16]

Answer:

Light moves at 300,000 kilometers per second, divide these and you get 500 seconds, or 8 minutes and 20 seconds this is an average number.

Explanation:

3 0
3 years ago
What is the minimum of acetic anhydride (102.1 g/mol) required to react completely with the 2.94 grams of salicylic acid? What v
Lady bird [3.3K]

Answer:

a) the minimun of acetic anhydride required for the  reaction is 2.175 g (CH3CO)2O

b) V acetic anhydride = 2.010 mL

Explanation:

  • balanced reaction:

      C6H4OHCOOH + (CH3CO)2O  ↔  C9H8O4 + C2H4O2

⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH

⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O /  mol  C6H4OHCOOH ) = 0.0213 mol (CHECO)2O

⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O

b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O

4 0
3 years ago
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