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Andreas93 [3]
3 years ago
9

There are 2.60 x 1015 short tons of oxygen in the atmosphere ( 1 short ton = 2000 lb). How many metric tons of oxygen are oresen

t (1 metric ton =1000 kg) ?
Chemistry
1 answer:
vichka [17]3 years ago
6 0

<u>Answer:</u> 2.36\times 10^{15} metric tons of oxygen is present.

<u>Explanation:</u>

We are given:

Mass of oxygen in atmosphere = 2.60\times 10^{15}\text{ short tons}

To convert it into metric tons, we use the conversion factors:

1\text{ short ton}=2000lb

And,

1\text{ metric ton}=2204.62lb

Converting short tons into pounds, we get:

\Rightarrow 2.60\times 10^{15}\text{ short tons}\times (\frac{2000lb}{\text{1 short ton}})=5.2\times 10^{18}lb

Now, converting the given mass in pounds to metric tons, we get:

\Rightarrow 5.2\times 10^{18}lb\times (\frac{\text{1 metric ton}}{2204.62lb})=2.36\times 10^{15}\text{metric ton}

Hence, 2.36\times 10^{15} metric tons of oxygen is present.

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Answer:

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3 0
3 years ago
What should i write on a essay about glue and liquid corn starch
PolarNik [594]
Hmm... interesting topic you're writing about here!
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8 0
3 years ago
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drek231 [11]

Answer:

27,397.23 L would be needed for a successful trip.

Explanation:

The problem gives us <u>the density (ρ) of the fuel,</u> by telling us that there are 803 g of fuel in 1 L, in which case:

ρ=\frac{mass}{Volume}=\frac{803g}{1L}  =803\frac{g}{L}

The required mass of fuel is 2.2 * 10⁴ kg, we can convert this value into g:

2.2 * 10⁴ kg * \frac{1000g}{1kg} = 2.2 * 10⁷ g

We calculate the required volume (V), using the mass and density:

803 g/L = \frac{2.2*10^{7}g }{V} \\V=\frac{2.2*10^{7}g }{803g/L}\\ V=27397.26 L

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4 years ago
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3 years ago
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Answer:

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Explanation:

I hope it helps you good luck

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