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amid [387]
3 years ago
13

Is ninhydren test positive for protein?

Chemistry
1 answer:
lyudmila [28]3 years ago
5 0
The Ninhydrin test is not effective to detect high molecular weight proteins as the steric hindrance limits the ninhydrin from reaching the α-amino groups.
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A ball has a volume of 9.0 liters and is subjected to a pressure of 3.5 atmospheres when its temperature is 74 °C. According to
vodka [1.7K]
To determine the number of moles of the gas, we need to relate moles to pressure, temperature and volume. To simplify things, we assume the gas is ideal so we use the ideal gas equation PV=nRT.

PV = nRT
n = PV / RT
n = 3.5 (9.0) / 0.08205 (74+273.15)
n = 1.1059 mol
6 0
4 years ago
What is the main function of starch in plants
galina1969 [7]
Explanation: it is formed by large number of glucose<span> molecules joined to each other , then when cell need it is digested to produce </span>glucose<span> to be consumed by the cell to produce energy.</span>
7 0
3 years ago
Read 2 more answers
Determine the molar mass of a 0.314-gram sample of gas having a volume of 1.6 l at 287 k and 0.92 atm. show your work.
Olenka [21]

As per Ideal gas equation, molar mass of the gas is 5.032 g/mo

We’ll begin by calculating the number of mole of the gas. This can be obtained as follow:

Volume (V) = 1.6 L

Temperature (T) = 287 K

Pressure (P) = 0.92 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

According to Ideal gas equation , PV = nRT

0.92 × 1.6 = n × 0.0821 × 287

1.472 = n × 23.5627

Divide by 23.5627

n = 1.176 / 23.5627

n = 0.0624 mole

Finally, we shall determine the molar mass of the gas. This can be obtained as follow:

Mass of gas = 0.314 g

Number of mole = 0.0624 mole

Mole = \frac{mass}{molar mass}

0.0624 = \frac{0.314}{molar mass}

Cross multiply

0.0624 × molar mass of gas = 0.314

Divide by 0.0624

Molar mass of gas =  \frac{0.314}{0.0624}

Molar mass of gas = 5.032 g/mo

Therefore the Molar mass of gas is 5.032 g/mo

Learn more about  Ideal gas equation here:

brainly.com/question/3637553

#SPJ1

8 0
2 years ago
What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure oc
White raven [17]

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says <em>"burn"</em>, it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

  • 55 L of gasoline (assuming gasoline to be pure octane).
  • The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the <em>conversión factor 1 L = 1000 cm³</em>.

55L.\frac{1000cm^{3} }{1L} =55000cm^{3}

Since <em>density = mass/volume</em>, we can solve for mass:

mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

  For octane:

  molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

  12 g/mol . 8 +  1g/mol . 18 = 114 g/mol

  According to the balanced equation reacts 1 mol of octane, which means      114 grams of it.

  For oxygen:

  According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is <u>114 g octane : 12.5 moles of oxygen</u>

<u />

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)

4 0
4 years ago
What would the name of the following compound be:
vodomira [7]

Answer:

ammonium phosphite

8 0
3 years ago
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