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3 years ago

Which has most likely been peer reviewed?

Chemistry

1 answer:

ElenaW [278]3 years ago

6 0

Answer:

an article on the beneficial bacteria in yogurt, featured in Medical Journal of Gastroenterology

Explanation:

This article is published in a Medical Journal, which means the contents must be credible, informative, and accurate. In order to be these things, the author most likely had others review their writing to remain objective and ensure quality work. The other articles are not meant to solely inform and are featured in less credible magazines. Therefore, the third option is correct.

You might be interested in

What is the molality of a solution containing 100 grams of glucose (C6H12O6, molar mass = 180 g/mol) dissolved in 2.5 kg of wate

meriva

Answer:

The answer to your question is molality = 0.22

Explanation:

Data

molality = ?

mass of glucose = 100 g

molar mass of glucose = 180 g

mass of water = 2.5 kg

Process

1.- Calculate the moles of glucose using proportions

180 g ------------------ 1 mol

100 g ------------------ x

x = (100 x 1) / 180

x = 100 / 180

x = 0.56 moles

2.- Calculate the molality

Formula

molality = moles / mass of solvent

- Substitution

molality = 0.56 / 2.5

- Simplification

molality = 0.22

6 0

3 years ago

Determine the percent composition of potassium dichromate, K2Cr2O7.

bulgar [2K]

Answer:

Option E is the correct one

Explanation:

K₂Cr₂O₇ → Potassium dichromate

A ionic salt which its molar mass is 294.18 g/m

It is composed by 2 mol of K, 2 mol of Cr and 7 mol of O

2 mol of K = 78.2 g

2 mol of Cr = 104 g

7 mol of O = 112 g

In 294.2 g of compound we have __78.2 g K __ 104 g Cr ___ 112 g O

Let's calculate the percent

(78.2 / 294.2) .100 = 26.59% K

(104 / 294.2) .100 = 35.35% Cr

(112 / 294.2) . 100 = 38.06 % O

4 0

3 years ago

The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN. Determine the average normal stre

rodikova [14]

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law,

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length,

ε = f / k

When the force is distributed between both materials will stretch the same length,

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

5 0

3 years ago

How many atoms are in 8 gram of helium

Yuri [45]

There are two atoms..............

5 0

3 years ago

A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre

Goryan [66]

Answer:

$\boxed {\boxed {\sf 82.7 \textdegree C}}$

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

$\frac {P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}$

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58 \ atm *2.46 \ m^3}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

$(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)$

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

$\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

$T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

The units of atmospheres and cubic meters cancel.

$T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}$

Solve inside the parentheses.

$T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}$

$T_2= \frac{369.246}{4.4625} \textdegree C}$

$T_2 = 82.74420168 \textdegree C$

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

$T_2 \approx 82.7 \textdegree C$

The temperature is approximately <u>82.7 degrees Celsius.</u>

3 0

3 years ago