Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983
Answer: 1.14
Explanation:

To find the molarity of the acid, we will utilize the equation derived from the neutralization process:

where
are the n-factor, molarity, and volume for the acid and
represent the n-factor, molarity, and volume for NaOH.
We know that:

By substituting the known values into the equation, we get:

To determine the pH of gastric juice:
The molarity amounts to = 0.072

Thus, the pH level of the gastric juice is 1.14
(c) Cu + S → CuS is classified as a redox reaction
Explanation:
The following reactions are presented:
(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl
(b) Pb²⁺ + 2 Br⁻ → PbBr₂
(c) Cu + S → CuS
Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:
Cu + S → CuS
In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.
Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).
Learn more about:
redox reactions
Response: 2. When molecules with different speeds collide, heat is transferred from the faster ones to the slower ones.
Clarification: I hope this was useful!:)
To find the answer, start by calculating the total mass of the copper utilized:
Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu
Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:
1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.
Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu
= 4.72 moles CuFeS2
The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole
CuFeS2 = 866.49 g CuFeS2