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Tanya [424]
3 years ago
5

What is the chemical formula for coper(ll)hydroxide?

Chemistry
1 answer:
kaheart [24]3 years ago
3 0
The chemical formula is Cu(OH)2
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A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,
Alja [10]

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

-Q= 66,300 J

-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

 Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J                117 J/0C x

------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

3 0
3 years ago
What are elementary particles the make up protons and nuetrons
ahrayia [7]

they are made by of two types of elementary particles

electrons and quarks

quetion answered by

 (jacemorris04)

4 0
3 years ago
Read 2 more answers
Lead has a density of 11.3 g/cm3. what is the volume of a 15.1 g block of lead?
Crank
Hey there:

density = 11.3 g/cm³

mass = 15,1 g

Therefore: 

D = m / V

11.3 = 15.1 / V

V = 15.1 / 11.3

V = 1.336 cm³
8 0
3 years ago
Draw the Lewis dot diagrams for F2, N2, and C2. Compare your drawing to the ones below. Which of the below Lewis dot diagram or
torisob [31]

answer: i believe it is N2 is wrong because it has a double bond

Explanation:

i need help as well

6 0
3 years ago
What is the expected freezing point of a 0.50 m solution of na2so4 in water kf for water is 1.86°c/m?
Anna007 [38]

Colligative properties calculations are used for this type of problem. Calculations are as follows:<span>


ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point)  = 0.93 °C
Tf - T = 0.93 °C
<span>T = -0.93 °C</span></span>

4 0
3 years ago
Read 2 more answers
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