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3 years ago

14

N2 H4 (1) + O2

1 answer:

Kruka [31]3 years ago

8 0

Answer:

d. 46.5 g

Explanation:

First, you need to start with a properly balanced equation:

N₂H₄(l) + 3O₂(g) -> 2NO₂(g) + 2H₂O(l)

Then, find the moles of water produced:

2H = 2.01568 amu

O = 15.999 amu

________________

18.01468 amu

52.4 g ÷ 18.01468 amu = 2.9087 moles

The mole ratio in our balanced formula between N₂H₄ : H₂O is 1 : 2, so divide moles of H₂O by 2 to get moles of N₂H₄

2.9087 moles ÷ 2 = 1.45435 moles of N₂H₄

Then, calculate the atomic mass of N₂H₄:

2N = 28.0134 amu

4H = 4.03136 amu

________________

32.04476 amu

Finally, calculate the mass in grams of N₂H₄:

1.45435 moles • 32.04476 amu = 46.604 g

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A partir de la desintegración del Virreinato del Río de la Plata y durante todo el siglo XXI, los nuevos Estados nacionales que

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Answer:

Verdadero (True).

Explanation:

Después de la desintegración del citado virreinato, los nuevos Estados intentaron establecer sus fronteras, a menudo a través de guerras e invasiones. Cabe destacar la invasión brasileña de Uruguay o la Guerra de la Triple Alianza. En menor medida, lo hicieron a través de tratados internacionales. (After the disolution of the viceroyalship described above, the new States attempted to establish their frontier usually by wars and invasions. It is to highlight the Brazilian invasion of Uruguay or the Triple Alliance' War. In a lesser extent, they made it through international treaties.)

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3 years ago

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2 years ago

Need it in 2 minutes ASAP

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3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<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>

3.77 mol K₃N

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

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

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3 years ago