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Zinaida [17]
4 years ago
15

Which process is an example of a physical change please help

Chemistry
1 answer:
Valentin [98]4 years ago
7 0
A physical change is one that a substance alters a substances form, but the substance has still not changed chemically. Carrots being cut up is changing the form of the carrots, but the fact that they are still carrots means it is a physical change. The answer is Carrots are cut into small pieces and mixed into a salad
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4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(ℓ) → 4 NaAu(CN)2(aq) + 4 NaOH(aq) (2) 2 NaAu(CN)2(aq) + Zn(s) → 2 Au(s) +Na2[Zn(CN)4](aq) I
marysya [2.9K]

Answer:

72.57 grams

Explanation:

The mass percentage of gold in the ore is 0.19 %

The mass of ore used = 23 kg

The mass of gold in the ore = \frac{0.19X23}{100}=0.0437kg

moles of gold in the given mass of ore =\frac{mass}{Atomic mass}=\frac{0.437X1000}{197}=2.22mol

As per given equation four moles of Au is giving four moles of complex compound on reacting with NaCN

Then in second reaction two moles of the complex is reacting with one mole of Zinc

Thus two moles of gold are reacting with one mole of Zinc

The moles of Zinc needed = 0.5 X 2.22 mol = 1.11 moles

The mass of Zinc needed = moles X atomic mass =1.11 X 65.38 = 72.57 grams

6 0
3 years ago
At a certain temperature and pressure, one liter of co2 gas weighs 1.55g. what is the mass of one liter of c2h6 gas at the same
finlep [7]
CO2:  12.0107 + 15.9994(2) = 44
NH3:   14.0067 + 1 (3) = 17


1.55 (17/44) = .599g
4 0
3 years ago
If an aqueous solution has a ph of 5.57 what is the concentration of hydroxide ion ([oh-]) in the solution?
irga5000 [103]
PH=5.57
pH+pOH=14
5.57 + pOH=14
pOH = 14-5.57= 8.43

pOH = - log  [OH⁻]
-pOH=  log  [OH⁻]


[OH^{-}] = 10^{-pOH}= 10^{-8.43}= 3.72*10^{-9}

 [OH^{-}]= 3.72*10^{-9}
8 0
3 years ago
What are three sources of information scientist use when they do research
exis [7]

Books, Other scientist's research, the Internet.

6 0
3 years ago
Name the following bases<br> -Ba(OH)2<br> - Ca(OH)2<br> -RbOH
Harman [31]

Ba(OH)_2\rightarrow \bold{\green{Barium\: hydroxide}}

Ca(OH)_2\rightarrow \bold{\red{Calcium\: hydroxide}}

RbOH \rightarrow \bold{\blue{Rubidium \:hydroxide}}

3 0
3 years ago
Read 2 more answers
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