What is the initial temperature of 50g of water that was raised to 50*C by the addition of 4.18kJ of heat energy? (2 sig fig)

The empirical formula of glucose is CH2O the molecular formula of glucose is 6 times than the empirical formula. what is the mol

jeka57 [31]

Answer:

The molecular formula of glucose is C₆H₁₂O₆

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of smallest whole number of atoms.

Molecular formula:

It gives the total number of atoms in a molecule of compound.

The molecular formula and empirical formula can be related as follow:

Molecular formula = n × empirical formula

Given data:

Empirical formula = CH₂O

Molecular formula = ?

It is stated in given problem that molecular formula is the 6 times of the empirical formula.

Molecular formula = n × empirical formula

Molecular formula = 6 × CH₂O

Molecular formula = C₆H₁₂O₆

The molecular formula of glucose is C₆H₁₂O₆.

5 0

3 years ago

compare a humans changes during exercise to the changes that occur a during a dogs exercise when it is exposed to a hot environm

svetoff [14.1K]

A human body, just like a dog's, will sweat. Dogs will pants and sweat through the pads of their feet to cool down, and human will sweat through their foreheads, armpits, etc.
Dogs will tend to, in hot environments, lay on the floor or where the surface is cooler. Since they cannot simply strip their clothing to keep cool they tend to find cool surfaces, fans, sources of air, etc. to keep cool from the heat.

5 0

3 years ago

Which weather technology can detect rain in tropical storms and hurricanes that are within about 250 miles, estimate how much ra

masha68 [24]

The answer is Doppler Radar. Hope this helps!

5 0

3 years ago

Susan conducts an experiment five times and gets a solution concentration of 1.9M, 2.1M, 1.8M, 1.9M, and 2.2M. The known concent

vlada-n [284]

Answer:

This question is incomplete; the complete part/options is:

A) They are precise, but not accurate.

B) They are accurate, but not precise.

C) They are both accurate and precise.

D) They are neither accurate nor precise

The answer is C.

Explanation:

Accuracy refers to how close a series of measurements is to a known or true value while precision is the closeness of the measurements to one another.

The known concentration of a solution is 2.0M. In an experiment which Susan conducts five times and gets a solution concentration of 1.9M, 2.1M, 1.8M, 1.9M, and 2.2M respectively.

Based on these measured solute concentrations, one would observe that the measured values of: 1.9M, 2.1M, 1.8M, 1.9M, 2.2M are close to one another. This means that the measurements are PRECISE.

Also, the measured values are not too distant from the known value of 2.0M, hence, it can be said that the measurements are ACCURATE.

Therefore, Susan's measurements of the solute concentration are both accurate and precise.

8 0

3 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 "burn", 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 conversión factor 1 L = 1000 cm³.

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

Since density = mass/volume, 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 114 g octane : 12.5 moles of oxygen

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

3 years ago