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Aleksandr [31]
3 years ago
11

Be sure to use proper significant figures. The molecular weight of NaCl is 58.44 grams/mole. To prepare a 0.50 M solution: You w

ould need to have grams of NaCl in 1,000 mL of water.\
Chemistry
2 answers:
Misha Larkins [42]3 years ago
5 0
Your answer would be,
Molarity = moles of solute/volume of solution we needed, 29.22(g)(mol) of NaCI

= 29.22(g)/58.44(g)(mol^-1)(1)/1(L)
= 0.500(mol)(L^-1)



Hope that helps!!!
Mice21 [21]3 years ago
4 0
The answer to the question is 29.22
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With the given chemical compounds, what is the balanced chemical equation when lit with fire?
Alchen [17]

Answer:

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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 <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 ion is formed when an atom of magnesium (mg) loses two electrons? mg–1 mg–2 mg 1 mg 2
frez [133]

the correct answer is Mg+2

3 0
3 years ago
Read 2 more answers
Balance this chemical equation. Choose "blank" for the box if no other coefficient is needed. Writing the symbol implies "1." H2
ololo11 [35]

4H202->2H2O+O2

Hope this helps mate good luck on whatever ur doing

6 0
3 years ago
Read 2 more answers
Gold cylinder has a mass of 75 g and a specific heat of 0.129J/G degrees Celsius it is heated to 65°C and then put in 500 g of w
nadezda [96]
<h3>Answer:</h3>

89.88° C

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of gold cylinder as 75 g
  • specific heat of gold is 0.129 J/g°C
  • Initial temperature of gold cylinder is 65°C
  • Mass of water is 500 g
  • Initial temperature of water is 90 °C

We are required to calculate the final temperature;

  • We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.
  • That is, Q = m × c × ΔT
<h3>Step 1: Calculate the quantity of heat absorbed by the Gold cylinder</h3>

Assuming the final temperature is X° C

Then; ΔT = (X-65)°C

Therefore;

Q = 75 g × 0.129 J/g°C × (X-65)°C

   = 9.675X - 628.875 Joules

<h3>Step 2: Calculate the quantity of heat released by water</h3>

Taking the final temperature as X° C

Change in temperature, ΔT = (90 - X)° C

Specific heat capacity of water is 4.184 J/g°C

Therefore;

Q = 500 g × 4.184 J/g°C × (90 - X)° C

  = 188,280 -2092X joules

<h3>Step 3: Calculate the final temperature, X°C</h3>

we know that the heat gained by gold cylinder is equal to the heat released by water.

9.675X - 628.875 Joules = 188,280 -2092X joules

2101.675 X = 188908.875

              X = 89.88° C

Thus, the final temperature is 89.88° C

3 0
3 years ago
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