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natka813 [3]
4 years ago
13

Hydrazine (N2H4) is a chemical compound used in rocket fuel. Which of the following represents the correct balanced equation for

the reaction of hydrazine liquid with oxygen gas to produce nitrogen gas and steam? A. N2H4 (l) + O (g) N (g) + H2O (g) B. N2H4 (l) + 2 O (g) 2 N (g) + 2 H2O (g) C. N2H4 (l) + O2 (g) N2 (g) + 2 H2O (g) D. N2H4 (l) + O2 (g) N2 (g) + H2O (g)
Chemistry
1 answer:
neonofarm [45]4 years ago
5 0


Answer is C.
N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g)

liquids should be denoted as (l) and gasses should be (g) including steam.
the natural form of oxygen gas and nitrogen gas are O₂ and N₂ respectively. 
Left hand side of equation:
N atoms = 2
H atoms = 4
O atoms = 2

Right hand side of equation:
N atoms = 2
H atoms = 2
O atoms = 1

hence the quation is imbalanced.
So 2 should be added at the left hand side of H₂O






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Anon25 [30]

Hey there!

The best way to balance chemical equations is to first start by balancing polyatomic ions such as OH and SO₄.

Next, balance other elements, but save elements that are by themselves for last, such as H₂ or Fe. Once you balance everything else you can do the ones by themselves, it's much easier.

Hope this helps!

6 0
3 years ago
A gas at 750 mmhg and with a volume of 2. 00 l is allowed to change its volume at constant temperature until the pressure is 600
Gemiola [76]

Answer:

The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.

Explanation:

Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;

P ∝ V

P V = K

P1 V1 = P2 V2

Parameters :

P1 = Initial pressure of the gas = 750 mmHg

V1 = Initial pressure of the gas = 2. 00 Liters

P2 = Final pressure of the gas = 600 mmHg

V2 = Fimal volume of the gas = ? Liters

Calculations :

V2 = P1 V1 ÷ P2

V2= 750 × 2. 00 ÷ 600

V2 = 1500 ÷ 600

V2 = 2.5 Liters.

Therefore, the new volume of the gas is 2. 5 Liters.

8 0
2 years ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
4 years ago
A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the weight
melomori [17]

Given:

volume of 0.08 m³

density of 7,840 kg/m³

Required:

force of gravity

Solution:

Find the mass using density equation.

D = M/V

M = DV

M = (7,840 kg/m³)(0.08 m³)

M = 627.2kg

 

F = Mg

F = (627.2kg)(9.8m/s2)

F = 6147N

7 0
3 years ago
What does the interquartile range represent?
Marizza181 [45]

Answer: C) middle 50 percent of the data

The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.

25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.

3 0
3 years ago
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