Answer: 37.5grams of Cu(NO3)2
Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2
<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>
<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>
<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>
<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>
<span>Answer:
1/4 is the average bond order for a pâ’o bond (such as the one shown in blue) in a phosphate ion.</span>
Answer: 0.9375 g
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of
solution = 0.75 M
Volume of
solution = 25.0 mL = 0.025 L
Putting values in equation 1, we get:
According to stoichiometry :
2 moles of
require = 1 mole of
Thus 0.01875 moles of
will require=
of
Mass of
Thus 0.9375 g of
is required to react with 25.0 ml of 0.75 M HCl
Answer:
this one is hard
Explanation:
but it's iron because the sodium so yea there u go.
Answer:
Aluminum nitrate is a salt composed of aluminum and nitric acid, belonging to a group of reactive chemicals - organic nitrate and nitrite compounds. The nitrate ion is polyatomic, meaning it is composed of two or more ions that are covalently bonded. This ion makes up the conjugate base of nitric acid.
Explanation: