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julia-pushkina [17]
2 years ago
11

A 29.0-g sample of water at 290. K is mixed with 51.0 g water at 300. K. Calculate the final temperature of the mixture assuming

no heat loss to the surroundings.
Chemistry
2 answers:
zubka84 [21]2 years ago
5 0
Let me conduct my research
Anvisha [2.4K]2 years ago
3 0

answer: 296 K

steps:

mass times temperature =mass times temperature

-51 * 300 = 29 * -290

300 * 51 = 15300

290 * 29 = 8410

-51x + 15300 = 29x - 8410

51 + 29 = 80

15300 + 8410 = 23710

23710 = 80x

80x = 23710

x = 296.375

https://socratic.org/questions/a-30-0-g-sample-of-water-at-280-k-is-mixed-with-50-0-g-of-water-at-330-k-how-wou

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How many moles are in 6.4grams of copper​
anzhelika [568]

Answer:

One mole of copper atoms has Avogadro number of copper atoms i.e. 6.022×10

23

 atoms.

Given the mass of Copper is 6.4g

Number of Copper atoms in 6.4g =

molar mass

given mass

​

×Avogadro number

                                                       =

63.55

6.4

​

×6.022×10

23

 copper atoms

                                                       =6.064×10

22

 copper atoms

Explanation:

8 0
3 years ago
a student carries out this chemical reaction in a beaker: Zn+CuCl2--&gt;Cu+ZnCl2 After the reactionhas come to completion, what
BigorU [14]

Answer:

The correct answer is

C. Cu and ZnCl2

Explanation:

I got it correct on PLATO

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If the PH is 5.85.how do I calculate for an acid ?[H^+] ​
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6 0
3 years ago
Production of Jam from Crushed Fruit in Two Stages. In a process producing jam (C1), crushed fruit containing 14 wt % soluble so
Andrej [43]

Answer:

The total kg from the mixer: m=2222.5 kg

Evaporated water: m_{water ev.}=791.1 kg

Jam produced: m_{jam}=1431.4 kg

Explanation:

Step by step:

1) Kg of mixture from the mixer

It says that crushed fruit is added to the mixer with the sugar and pectin.

The fruit added is 1000 kg

The sugar added is 1.22 kg sugar/1 kg crushed fruit:

m_{sugar}=1000 kg fruit * \frac{1.22 kg sugar}{1 kg fruit}=1220 kg sugar

The pectin added is 0.0025 kg pectin/1 kg crushed fruit:

m_{pectin}=1000 kg fruit * \frac{0.0025 kg pectin}{1 kg fruit}=2.5 kg pectin

The total kg from the mixer:

m=1000 kg fruit + 2.5 kg pectin + 1220 kg sugar=2222.5 kg

2) Evaporated water

To calculate the evaporated water it's important to have in mind that the mix goes from a concentration of 14 wt% to 67 wt%. This difference is because of the evaporated water. So:

Initial: 1000 kg of fruit with 14 wt% of solids

m_{solids}=1000 kg *0.14=140 kg

<em>This amount of solids is constant </em>

Final: The mass of solids is the same but now it represents the 67 wt%

m_{final}=140 kg *\frac{100}{67}=208.9 kg

and

m_{water ev.}=1000kg-m_{final}

m_{water ev.}=1000kg-208.9kg=791.1 kg

3) Jam produced

m_{jam}=m_{final}+m_{pectin}+m_{sugar}=208.9 kg + 1220 kg + 2.5 kg

m_{jam}=1431.4 kg

6 0
3 years ago
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