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Hitman42 [59]
3 years ago
12

Why is peer review important when publishing experimental results? A. Peer review makes the experiment repeatable. B. Peer revie

w is necessary for the scientific method to be followed. C. Peer review keeps the data from being biased. D. Peer review makes the results more reliable.
Chemistry
2 answers:
MA_775_DIABLO [31]3 years ago
4 0
The answer is D. At least that's what I think.
cluponka [151]3 years ago
3 0

Answer:

D. Peer review makes the results more reliable.

Explanation:

Hello,

When one publishes experimental results, during the experimentations, several mistakes such as wrong data record, fluctuating-displayed measurements (specially by the balances), trembling while pipetting, wrong procedures and others could be done so the results will be affected. Thus, a peer is important because that person could help us to correct things we're doing wrongly or do them again to get a common agreement between the employed techniques and recordings.

Best regards.

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A- law of conservation of energy
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What type of model did Rutherford, Thompson and Bohr develop?
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The Bohr atomic model, relying on quantum mechanics, built upon the Rutherford model to explain the orbits of electrons.
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1. Which has the greatest heat capacity?
mash [69]

Answer:

as the greatest heat capacity? a. 1,000 g of water b. 1,000 g of steel c. 1 g of water d. 1 g ... +1. kvargli6h and 1 other learned from this answer. Answer: a. 1,000 g of water ... Heat capacity of steel = 0.49 J/gram^0C. Hence 1,000 g of water will have greatest heat capacity.

Explanation:

4 0
3 years ago
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th
satela [25.4K]

Answer:

General equation for a double-displacement reaction:  

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄  

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄  

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄  

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃  

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃  

NaOH + AgNO₃   --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃  --> NaNO₃ +  Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0
3 years ago
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