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tino4ka555 [31]

2 years ago

10

Which of the following would be most useful in trying to obtain procedural information to replicate an experiment previously pub

lished?

1 answer:

Jet001 [13]2 years ago

7 0

Answer: Peer-reviewed journal article is the most useful because the information in them had been carefully scrutinized and aproved by people who are experts in that particular field.

You might be interested in

How many O atoms are in 4C₆H₁₂O₆

ipn [44]

Number of O atoms : 24

<h3>Further explanation</h3>

Given

C₆H₁₂O₆ compound

Required

Number of atoms

Solution

A molecular formula shows the number of atomic elements in compound.

The empirical formula is the smallest comparison of the atoms

Glucose-C₆H₁₂O₆ is composed of 3 elements, namely C, H, and O.

The number of atoms in a compound can usually be seen from the subscript number after the atom and the reaction coefficient shows the number of molecules

So number of O atoms :

= 4 x 6 = 24 atoms

6 0

2 years ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

2 years ago

Which best compares a molecule that has a trigonal planar shape with one that has a trigonal pyramidal shape?

Juli2301 [7.4K]

Answer:

B. They both contain three atoms around the central atom.

Explanation:

Do the unit test on edg20 and got it right!

7 0

3 years ago

What is volume? What instrument is used to measure liquid volume? What formula is used to calculate the volume of a solid object

sertanlavr [38]

Answer:

Volume is the amount of space that a substance occupies. A graduated cylinder is used to measure liquid volume. The formula used to calculate the volume of a solid object is = L x W x H

Explanation:

8 0

2 years ago

El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si

Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

$PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)$

Lo cual hace que la expresión de equilibrio se calcule como:

$Ksp=[Pb^{2+}][Cl^-]^2$

Y que en términos de la solubilidad molar, s, se resuelve como:

$1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L$

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

$s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L$

¡Saludos!

7 0

2 years ago