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kirill [66]
3 years ago
14

1. A 40,000 kg plane is accelerating at a rate of 4m/s2. How much force does the engine put forth?​

Chemistry
1 answer:
mamaluj [8]3 years ago
5 0

Answer:

<em>160000</em>

Explanation:

F=ma

m=40000

a=4

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A Series of choices between two characteristics that is used to identify an organism is called a
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Answer: Dichotomous

Using a Dichotomous Key to Identify Organisms

A dichotomous key is a tool used by biologists to identify organisms in a group through a process of answering yes or no questions about the organism. Dichotomous means 'divided into two parts', as each question about an organism has two choices.

8 0
4 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0
3 years ago
Select the pair that has the larger atom or ion listed first.
Sedbober [7]

Answer:

Correc option: Br^- \, , Kr

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size of atom : it says somthing about how many shell present in a particular atom or ion and it can also be evaluated on the basis of radius of atom.

Br^- and Kr has highest number of shell as compared to other group of species .

Na ,S , Mg ,P all are from 3rd period but Kr and Br^- in the 4th period so size of species of this group will more,

Size increases on increasring the shell number

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The answer is A :) your welcome hope this helps
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