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qaws [65]
2 years ago
7

A jet airplane reaches 750 km/h on a certain flight. What distance does it cover in 10 min?

Chemistry
2 answers:
malfutka [58]2 years ago
3 0

The distance that jet covered in 10 min is 125 Km

calculation

distance = speed  x time

speed=750 km/h

time= 10/60hrs

=750 x10/60=125 Km

telo118 [61]2 years ago
3 0

Answer : The distance covered by jet airplane is 125.25 km

Explanation :

Speed : It is defined as the distance traveled by the object per unit time.

Formula used :

Speed=\frac{Distance}{Time}

Given:

Speed of jet airplane = 750 km/hr

Time = 10 min = 0.167 hr

conversion used : (1 hr = 60 min)

Now we have to determine the distance covered by jet airplane.

750km/hr=\frac{Distance}{0.167hr}

Distance=125.25km

Therefore, the distance covered by jet airplane is 125.25 km

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A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate
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Answer:

0.08 mol L-1

Explanation:

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Ammonia Formula: NH3

Ammonium sulfate Formula: (NH₄)₂SO₄

H2SO4 + 2NH3 = 2NH4+ + SO4 2-

H2SO4 + 2NH3 = (NH₄)₂SO₄

H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L

= 0.08 mol L-1

7 0
2 years ago
What is the boiling point, melting point, and triple point of bromine?
Lapatulllka [165]

Answer:

A. Boiling point = 59 °C, Melting point = -7.2°C, triple point = -7.3°C

Explanation:

3 0
3 years ago
Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g
Agata [3.3K]
An exponential decay law has the general form: A = Ao * e ^ (-kt) =>

A/Ao = e^(-kt)

Half-life time => A/Ao = 1/2, and t = 4.5 min

=> 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154

Now replace the value of k, Ao = 28g  and t = 7 min to find how many grams of Thalium-207 will remain:

A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g

Answer 9.5 g.
7 0
3 years ago
The most common cooling mechanism for cloud formation is ________.
KatRina [158]
The answer is "Rising and expanding air or atmosphere cooling".
To form a cloud the air must be cooled to the temperature of dew point. when there is expansion of air, it cools to the dew point and thus the formation of cloud happens. This process is very common and used for the formation of clouds.
8 0
2 years ago
2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g
julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system.  Now;

Number of moles of Fe =  12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with  0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of  Fe3O4(s)

 0.22 moles  of O2   produces   0.22 moles * 1 mole/3 moles of  Fe3O4(s) = 0.1073 moles

Mass of  Fe3O4(s) =0.1073 moles  * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0
2 years ago
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