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guajiro [1.7K]
3 years ago
13

What do greenhouse gases do in our atmosphere?

Chemistry
1 answer:
Vitek1552 [10]3 years ago
3 0

Explanation:

Greenhouse gases are gases in Earth's atmosphere that trap heat. They let sunlight pass through the atmosphere, but they prevent the heat that the sunlight brings from leaving the atmosphere.

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What is the velosity of a 72.3 kg jogger with a kinetic energy of 1080.0 J?
Svet_ta [14]

Answer: 5.47m/s

Explanation:

Mass = 72.3kg

K.E = 1080.0J

V =?

K.E = 1 /2MV^2

V^2 = 2K.E /M = (2x1080)/72.3

V = sqrt [(2x1080)/72.3]

V = 5.47m/s

7 0
3 years ago
For the following word equations, write it as a chemical equation, then balance it.
Lorico [155]

4K+O2-----------2K2O

7 0
3 years ago
Element X is more reactive than lithium and magnesium but less reactive than potassium. Which element is most likely to be X?
PilotLPTM [1.2K]

Answer:

It is Likely to Be Sodium (Na) coz as You Down The group the reactivity increase

3 0
3 years ago
How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
AVprozaik [17]
MThe  heat  energy  required  to  raise  the  temperature   of  0.36Kg   of  copper   from   22 c   to  60  c  is  calculate  using  the  following  formula

MC delta T
m(mass)=  0.360kg  in  grams  =  0.360  x1000 = 360 g
  c(specific  heat  energy)  =  0.0920  cal/g.c
delta T =  60- 23  = 37  c

heat  energy is therefore=  360g   x0.0920 cal/g.c  x 37  c=  1225.44  cal

5 0
3 years ago
An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had
Vlad [161]

Answer : The time passed in years is 2.74\times 10^2\text{ years}

Explanation :

Half-life of carbon-14 = 5730 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant disintegrate = 15.3

a - x = amount left after decay process = 14.8

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{15.3}{14.8}

t=274.64\text{ years}=2.74\times 10^2\text{ years}

Therefore, the time passed in years is 2.74\times 10^2\text{ years}

4 0
4 years ago
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