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amid [387]
3 years ago
7

I NEED THE ANSWER ASAP! Its science btw

Chemistry
1 answer:
Roman55 [17]3 years ago
6 0
A. Crystals of minerals dissolve in the groundwater in caves
You might be interested in
How many eggs are in 1 mole of eggs?
Pachacha [2.7K]

Answer:

I <em>believe</em> it is Avogadro's number: 6.022 × 10²³

Explanation:

One mole of a substance is equal to 6.022 × 10²³ units of that substance

4 0
3 years ago
The Statue of Liberty is 111.5 feet tall. What would be the mass in KG of a stack of quarters equal her height? A quarter is 1.7
miskamm [114]

Answer:

( About ) 110 kilograms

Explanation:

Take a look at the attachment below,

6 0
4 years ago
If there is 25.0 L of gas at a temperature of 350C what is the temperature if the volume increases to 75.0L?
Zielflug [23.3K]

Answer:

The new temperature is 1596 C.

Explanation:

Charles's Law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.

Statistically, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio that exists between the volume and the temperature will always have the same value:

\frac{V}{T}=k

Being an initial state 1 and a final state 2, it is fulfilled:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 25 L
  • T1= 350 C= 623 K (being 0 C= 273 K)
  • V2= 75 L
  • T2= ?

Replacing:

\frac{25 L}{623 K} =\frac{75 L}{T2}

Solving:

T2*\frac{25 L}{623 K} =75 L

T2=75 L*\frac{623 K}{25 L}

T2= 1869 K= 1596 C

<u><em>The new temperature is 1596 C.</em></u>

4 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 <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
Balance the following equations:
Brilliant_brown [7]

Answer:

1) 0 C2H4O2 + 0 O2 -> 0 CO2 + 0 H2O (balanced)

2) V2O5 + CaS -> CaO + V2S5

<em>just additional info: V2O5 </em><em>is</em><em> </em><em>divanadium</em><em> </em><em>pentaoxide</em>

LHS (Left hand side)

V: 2

O: 5

Ca: 1

S: 1 x 5 [to balance with the right hand side of the equation]

RHS (Right hand side)

V: 2

O: 1 x 5 [to balance with the left hand side of the equation]

Ca: 1

S: 5

When you balance any elements, you have to balance the whole chemical compound.

Thus,

V2O5 + <em><u>5</u></em> CaS -> <em><u>5</u> CaO</em> + V2S5

LHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

RHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

3) S8 + O2 -> SO2

LHS:

S: 8

O: 2

RHS:

S: 1 x 8 [to balance with LHS]

O: 2

When you balance any elements, you have to balance the whole chemical compound.

S8 + O2 -> <em><u>8</u></em><em><u> </u></em>SO2

When we add 8 to the RHS, it gives us 8S, 16 O.

In order to balance that into the RHS, I need to multiply the O2 by 8, which will give 8(O2) = 16 O particles.

Therefore, <em><u>S8</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>O2</u></em><em><u> </u></em><em><u>-</u></em><em><u>></u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>SO2</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>final</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>(</u></em><em><u>3</u></em><em><u>)</u></em><em><u>.</u></em>

6 0
3 years ago
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