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3 years ago

5

For the complete combustion of 1.000 mole of methane gas at 298 K and 1 atm pressure, ΔH° = -890.4 kJ/mol. What will be the amou

nt of heat released when 1.97 g of methane is combusted under these conditions?

1 answer:

anastassius [24]3 years ago

3 0

Oh hooray a practice problem:

That might be wrong

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Please help!!!!!! BRAINLIEST to right answer! :))

notsponge [240]

Explanation:

d. endothermic change as

heat is added to solid ice to change it to liquid water

6 0

3 years ago

Please help me get the answer to bothe a and b

neonofarm [45]

Answer:

If a metal and metal solution react, the more reactive metal will displace the less reactive metal from solution. If the metal in solution you start with is formed from a more reactive metal than the metal to be added, no reaction will occur.

3 0

2 years ago

I can't find the answer for number 9/10 pleas help

kondaur [170]

(10)an object will continue to travel at a constant speed unless acted by an unbalanced force according to newtons second law every force acted on a body has a equal and opposite reaction so the speed and the direction of the object will change.

(9) balanced force (i think so )

hope this is what you needed the 10th one is for understanding you can shorten it after reading and i think what i have written is not wrong :P:P:P;p:P:):):):D=) for the ninth one i am not sure but i think so :P:)=)

8 0

3 years ago

Read 2 more answers

3. Select the one which is neither an acid nor a base. *

irinina [24]

since, a,b is an acid.. imposibble it is the answer

we left with c and d

while we can see that 'hyroxide' is a base

answer: C

3 0

3 years ago

If charge on only one object is tripled, determine the new force between them:

Allisa [31]

Answer:

+3·F

Explanation:

The number of objects in the given system = 2 objects

The charge on each object are; q₁ = -Q, q₂ = -Q

The force acting between the objects = +F

The distance between the objects = 2·d

The formula for the force acting between two charged particles is given as follows;

$F=K \times \dfrac{q_{1} \times q_{2}}{r^{2}}$

Therefore, we get;

$F=K \times \dfrac{-Q \times -Q}{(2\cdot d)^{2}} = K \times \dfrac{Q^2}{4 \cdot d^2}$

By tripling the charge, q₁, on the first object, we get;

q₂ = 3 × (-Q)

$F_2=K \times \dfrac{-3 \cdot Q \times -Q}{(2\cdot d)^{2}} = K \times \dfrac{3 \cdot Q^2}{4 \cdot d^2} = 3 \times +F = +3\cdot F$

Therefore, the new force between them, F₂ = +3·F

5 0

3 years ago