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KengaRu [80]
2 years ago
8

Student Z completes his/her calibration step for part 3 and finds an average massA to be 95.237 g ± 0.005 g. However, in his/her

calculations, s/he mistakenly uses 85.237 g ± 0.005 g for massA. S/he uses the correct values for massB and masscyl. Will the calculated mass of displaced water be too high, too low, or unchanged? And why?
Chemistry
2 answers:
Troyanec [42]2 years ago
7 0
It is going to be too low because the mass mistakenly used is lower than the initial.
cupoosta [38]2 years ago
5 0

Answer:- Calculated mass of displaced water would be too low.

Explanations:- Mass of water displaced is directly proportional to the mass of the object dropped into water.

Since the average mass of A used for calculations is low than the actual mass, the calculated mass of displaced water would also be low.

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Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder. What is the formul
Mars2501 [29]

Answer:

Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).

Explanation:

This is a chemical reaction problem.

In expressing any chemical reaction, we need to understand that there are reactants and products.

  • The reactants are the species on the left hand side that are combining.
  • The products are the species on the right hand side that are formed.
  • Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.

Using the statement of this problem, we can deduce that;

 Reactants are Fluorine gas and Calcium metal

  Product is Calcium Fluoride

Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.

                    F₂_{(g)}     +    Ca_{(s)}               →           CaF₂_{(s)}

We can see that equal number of atoms are on both sides of the expression.

7 0
2 years ago
Read 2 more answers
Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. a
kirill115 [55]
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?

m(Mg)=wm
m(Zn)=(1-w)m

Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)

Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)

m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)

w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))

w=0.583 (58.3%)
5 0
2 years ago
Acidosis is a human body disorder caused by too much acid in the stomach. If a patient was tested for this using their saliva an
Luden [163]

it will turn green or blue I think

8 0
2 years ago
Rank the formation of the solutions A, B, and C from the most exothermic to the most endothermic. Rank the enthalpy of solution
Elodia [21]

This is an incomplete question, the table is attached below.

Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

Explanation :

As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.

In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.

In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.

Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

4 0
2 years ago
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
2 years ago
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