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sergejj [24]
2 years ago
11

Which part of experimental design is most important to a scientist when replicating an experiment? Having exactly the same data

as the previous experiment using the same procedure and variables as the previous experiment conducting an experiment similar to the previous experiment using the same laboratory that was used in the previous experiment
Chemistry
2 answers:
yarga [219]2 years ago
5 0

Answer:

The answer to your question is below:

Explanation:

Having exactly the same data as the previous experiment I think that having the same data as the previous experiment is extremely important but not the most important, for me is the second most important.

Using the same procedure and variables as the previous experiment For me, this is the most importan thing when a scientist is designing an experiment, because if he or she follow exactly the same procedure and variables, then the results will be very close.

Conducting an experiment similar to the previous experiment  This characteristic is important but not the most important.

Using the same laboratory that was used in the previous experiment It is not important the laboratory, if the procedure and variables are the same, your experiment must give the same results in whatever laboratory.

Elodia [21]2 years ago
4 0

Answer:

B) Using the same procedure and variables as the previous experiment

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6.25 g of nickel -63 will be left

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Answer:

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Explanation:

hypothesis is a scientific guess made be scientist in order to give account for the natural phenomenon

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2 years ago
The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.
xxMikexx [17]

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

6 0
2 years ago
The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr
masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is C_7H_{11}O_2SN

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

Mass of SO_2=0.4503g

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, \frac{12}{44}\times 2.1654=0.590g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, \frac{2}{18}\times 0.6965=0.077g of hydrogen will be contained.

  • <u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, \frac{32}{64}\times 0.4503=0.225g of sulfur will be contained.

  • <u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, \frac{14}{30}\times 0.2109=0.098g of nitrogen will be contained.

  • Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles

Moles of Sulfur = \frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = \frac{0.049}{0.007}=7

For Hydrogen  = \frac{0.077}{0.007}=11

For Oxygen  = \frac{0.0142}{0.007}=2.03\approx 2

For Sulfur  = \frac{0.007}{0.007}=1

For Nitrogen  = \frac{0.007}{0.007}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN

4 0
2 years ago
Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0
2 years ago
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