Answer:
300 gr.
Explanation:
if 204gr/100gr=solubility, then according to the condition solubility=612/x.
It means, 204/100=612/x, where x=612*100/204=300 gr.
I need help ASAP
1.Minerals are generally (2 points)
A.solids
B.liquids
C.gases
D.plasmas
2.Which of these statements is correct about water and oil? (2 points)
A.Water is not a mineral because it is a liquid but oil is a mineral because it is formed under the ground by natural processes.
B.Both water and oil are minerals because they are natural materials with a definite chemical composition.
C.Water is a mineral because it has a definite chemical composition but oil is not a mineral because it is organic.
D.Both water and oil are not minerals because they are liquids
3.What mineral is most likely used to make the plumbing pipe? (2 points)
A.talc
B.zinc
C.quartz
D.copper
4.Which of these is the correct next step to complete the diagram? (2 points)
A.Put an arrow labeled cools and crystallizes, pointing from igneous rocks to sediments.
B.Put an arrow labeled cools and crystallizes, pointing from sedimentary rocks to sediments.
C.Put an arrow labeled heat and pressure, pointing from igneous rocks to metamorphic rocks.
D.Put an arrow labeled heat and pressure, pointing from metamorphic rocks to sedimentary rocks.
5.
Igneous rocks are formed due to the (2 points)
A..cooling of magma
B.weathering of rocks
C.changes in pressure
D.changes in temperature
6.David plans to grow corn on his farm. Corn grows best in clay loam soil. Which of these is the most likely soil sample that David would select to grow corn? (2 points)
A.sample containing 30 percent sand, 35 percent clay, and 35 percent silt
B.sample containing 25 percent sand, 15 percent clay, and 60 percent silt
C.sample containing 40 percent sand, 15 percent clay, and 45 percent silt
D.sample containing 55 percent sand, 30 percent clay, and 15 percent silt
1. a
2. d
3. d
4. c
5. a
6. a
Answer:
A.solids
D.Both water and oil are not minerals because they are liquids
D.copper
C.Put an arrow labeled heat and pressure, pointing from igneous rocks to metamorphic rocks.
A..cooling of magma
A.sample containing 30 percent sand, 35 percent clay, and 35 percent silt
What is the molar concentration of the acid if 35.18 mL of hydrochloric acid was required to neutralize 0.745 g of ALUMINUM hydroxide?
The solution will be 0.27M hydrochloric acid
In the final chemical equation, NaCl and O2 are the products that are formed through the reaction between Na2O and Cl2. Before you can add these intermediate chemical equations, you need to alter them by
the answer to the question is b
Answer : The correct option is, (B) multiplying the equation first by 2.
Explanation :
The balanced final chemical reaction will be,
[tex]2Na_2O(s)+2Cl_2(g)\rightarrow 4NaCl(s)+O_2(g)[/tex]
The two intermediate chemical reaction are give :
(1) [tex]2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)[/tex]
(2) [tex]2Na_2O(s)\rightarrow 4Na(s)+O_2(g)[/tex]
First multiplying the equation (1) by 2, we get
[tex]4Na(s)+2Cl_2(g)\rightarrow 4NaCl(s)[/tex]
Now adding both the equations, we get
[tex]2Na_2O(s)+4Na(s)+2Cl_2(g)\rightarrow 4Na(s)+O_2(g)+4NaCl(s)[/tex]
The final chemical reaction are :
[tex]2Na_2O(s)+2Cl_2(g)\rightarrow 4NaCl(s)+O_2(g)[/tex]
Hence, the correct option is, (B) multiplying the equation first by 2.
Chalcopyrite, the principal ore of copper (Cu), contains 34.63 percent Cu by mass. How many atoms of Cu can be obtained from 5.11 x 103 kg of ore? A) 1.68 x 1028 Cu atoms B) 1.68 x 1025 Cu atoms C) 6.77 x 1031 Cu atoms D) 6.77 x 1034 Cu atoms
Answer:
A) 1.68 x 10²⁸ Cu atoms.
Explanation:
∵ The mass of Cu in the chalcopyrite ore = mass of the ore x weight fraction of Cu.
Weight fraction of Cu = Weight % / 100 = 34.63 /100 = 0.3463.∴ The mass of Cu in the chalcopyrite ore = mass of the ore x weight fraction of Cu = (5.11 x 10³ kg)(0.3463) = 1.77 x 10³ kg.
We need to calculate the no. of moles of Cu atoms:n of Cu = mass/atomic mass of Cu = (1770 x 10³ g) / (63.54 g/mol) = 27.853 x 10³ mol.
We have that every mole of an element contains Avogadro's no. of atoms (6.022 x 10²³).∴ 27.853 x 10³ mol of Cu contains = (27.853 x 10³ mol)(6.022 x 10²³) = 1.677 x 10²⁸ Cu atoms ≅ 1.68 x 10²⁸ Cu atoms.