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3 years ago

Lucy wants to attach a goal cost to each of her life goals. Why might she do this?

Chemistry

1 answer:

Natasha2012 [34]3 years ago

8 0

Answer:

So she can have something to reach or look forward to.

Explanation:

none

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Gasoline has a density of 0.749 g/mL. How many pounds does 19.2 gallons of gasoline weigh? Use significant figures. Do not enter

Yanka [14]

In this question, you are given the gasoline density (0.749g/ml) and volume of the gasoline (19.2 gallons). You are asked the mass of the gasoline in pounds. Then you need to change the grams into pounds and the ml into gallons. The calculation would be:

mass of gasoline= density * volume
mass of gasoline= 0.749g/ml * (1 pound/453.592grams) * 3785.41ml/gallon * 19.2 gallon= 120 pounds

7 0

3 years ago

What is the relationship between the amount of carbon dioxide and the earths average temperature?

Anit [1.1K]

Answer:

carbon dioxide concentration goes down, temperature goes down. Carbon dioxide goes up temp goes up, carbon dioxide is directly related to temperature by insulating it in the earths atmosphere and if there's less of it then the temp would go down.

Explanation:

6 0

2 years ago

Exactly 25.0 mL of an aqueous solution of barium hydroxide required 32.0 mL of 0.200 M nitric acid to neutralize it. What is the

Brrunno [24]

Answer:

2000m the example 25.0\0.200

7 0

3 years ago

Read 2 more answers

A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C

Mamont248 [21]

Answer:

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

6 0

3 years ago

Read 2 more answers

Certain hydrocarbon is 92.3 % carbon and 7.7 % hydrogen by mass.If the molar mass of the hydrocarbon is approximately 40 g/mol,

zimovet [89]

Answer:

The hydrocarbon has a molecular formula of C3H3

Explanation:

Step 1: Data given

Suppose the mass of the compound is 100 grams

A hydrocarbon contains:

⇒ 92.3 % carbon = 92.3 grams

⇒ 7.7 % hydrogen = 7.7 grams

Molar mass of the compound is 40 g/mol

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles carbon = 92.3 grams / 12.01 g/mol

Moles carbon = 7.69 moles

Moles hydrogen = 7.70 grams / 1.01 g/mol

Moles hydrogen = 7.62

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

Carbon: 7.69/7.62 = 1

Hydrogen: 7.62/7.62 = 1

The empirical formula is CH

The molar mass is 13.02 g/mol

Step 4: Calculate the molecular formula

n = 40 g/mol / 13.02 g/mol

n ≈ 3

We need to multiply the empirical formula by 3 to get the molecular formula

Molecular formula = 3*(CH) = C3H3

The hydrocarbon is C3H3

5 0

3 years ago