1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
koban [17]
3 years ago
7

A lab group made a copper amine stock solution with a concentration of 0.02 M. A group member transfers 9.27 mL of this solution

to a 25 mL volumetric flask and dilutes to the line with water. What is the concentration of the resulting solution?
Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer : The concentration of the resulting solution is, 0.0074 M

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of stock solution.

M_2\text{ and }V_2 are the final molarity and volume of diluted solution.

We are given:

M_1=0.02M\\V_1=9.27mL\\M_2=?\\V_2=25mL

Putting values in above equation, we get:

0.02M\times 9.27mL=M_2\times 25mL\\\\M_2=0.0074M

Hence, the concentration of the resulting solution is, 0.0074 M

You might be interested in
Can you please help me !!
ioda

Answer:

Explanation:

A)  O2 (non polar covalent)

B)  HF  (polar covalent)

C) NaCl (because its ionic)

im not 100% sure hope it helps

7 0
3 years ago
A light-year is a way to measure ______ in space.
kirill115 [55]

Answer:

Distance

Explanation:

The light-year is a measure of distance, not time. It is the total distance that a beam of light, moving in a straight line, travels in one year.

4 0
3 years ago
Read 2 more answers
Milk of magnesia, a suspension of mg(oh)2 in water, reacts with stomach acid (hcl) in a neutralization reaction. mg(oh)2(s) + 2
defon
Balanced equation for the above reaction is as follows;
Mg(OH)₂ + 2HCl ---> MgCl₂ + 2H₂O
stoichiometry of Mg(OH)₂ to MgCl₂ is 1:1
mass of Mg(OH)₂ reacted - 1.82 g
number of moles of Mg(OH)₂  - 1.82 g/ 58.3 g/mol = 0.0312 mol
number of Mg(OH)₂  moles reacted - number of MgCl₂ moles formed 
number of MgCl₂ moles formed - 0.0312 mol
mass of MgCl₂ formed - 0.0312 mol x 95.2 g/mol = 2.97 g
mass of MgCl₂ formed - 2.97 g

5 0
4 years ago
Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a
Crank

Explanation:

a. Pb(NO_3)_2(aq) + Na_2SO_4(aq) → ?

Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)

Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)

Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)

Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)

b. NiCl_2(aq) + NH_4NO_3(aq) →

NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)

No precipitation is occuring.

c. Fe_Cl2(aq) + Na_2S(aq) →

FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)

FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)

Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)

Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)

d.MgSO_4(aq) + BaCl_2(aq) →

MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2

MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)

BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)

Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

5 0
3 years ago
Number of grams of hydrogen than can be prepared from 6.80g of aluminum​
Anastaziya [24]

Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

5 0
4 years ago
Other questions:
  • According to Dalton's atomic theory, which statement is true about atoms of the same element? They can be destroyed. They have i
    8·2 answers
  • PLEASE HELP I NEED IT!!!!!
    15·1 answer
  • What is milk best described as? solution, suspension, compound, or colloid?
    12·1 answer
  • Which element has a larger atomic radius than sulfur?
    8·1 answer
  • Which of the following characteristics of the pineal gland is not correct? A. The pineal gland is located in the brain, above th
    11·1 answer
  • How many carbon atoms are there in a 1.5-carat diamond? diamonds are a form of pure carbon. ( 1 carat=0.20 grams)?
    15·1 answer
  • Is a CAS number a physical or chemical property
    6·1 answer
  • What accommodations have been designed to protect astronauts from micrometeoroids hitting their spacesuits?
    9·1 answer
  • How is water used in society
    12·2 answers
  • Consider the reaction.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!