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3 years ago

35000 kJ/ 0.250 s kJ is kilojoules

1 answer:

5 0

The unit 'kJ' or kilojoules is a unit of measurement for energy. It is equivalent to the SI units kg·m²/s². When energy is divided with time, that unit is now in terms of work, which energy per unit time. The parameter kJ/s is equivalent to kW or kilowatts.

Work = 35,000 kJ/0.250 s = <em>140,000 kW</em>

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Find density of nitrogen dioxide at 75*C and 0.805 atm.

Eva8 [605]

Answer:

1 (348) (D2) = 273 (2.05) (0.805) D2= 1.29 g/L

Explanation:

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4 years ago

What type of chemical reaction is oil spills?

labwork [276]

Decomposition, because the oil eventually breaks down bc of bacterias and evaporate into the ocean!

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2 years ago

What is the change in boiling point for a 0.615m solution of Mgl2 in water?

ivanzaharov [21]

Answer : The change in boiling point is, $0.94^oC$

Explanation :

Formula used :

$\Delta T_b=i\times K_f\times m$

where,

$\Delta T_b$ = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

$K_f$ = boiling point constant for water = $0.51^oC/m$

m = molality = 0.615 m

Now put all the given values in this formula, we get

$\Delta T_b=3\times (0.51^oC/m)\times 0.615m$

$\Delta T_b=0.94^oC$

Therefore, the change in boiling point is, $0.94^oC$

4 0

3 years ago

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

4 0

4 years ago

How many moles of S would I have if I had 11 grams? (Stoichiometry) HELP

zepelin [54]

<h3>Answer:</h3>

0.34 mol S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

11 g S

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 11 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})$
Multiply/Divide: $\displaystyle 0.343 \ mol \ S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.343 mol S ≈ 0.34 mol S

4 0

3 years ago