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HACTEHA [7]
3 years ago
8

BRAINLIEST ANSWER! PLEASE HELP!! An Earth-like planet is 500 light years away. Why is this an obstacle for a manned space missio

n?
A. The cost is too low
B. There is no oxygen in space
C. Humans cannot live for 500 years
D. There are no plans to send a human
Chemistry
1 answer:
Setler79 [48]3 years ago
7 0
I Would think it would be C Since Humans can't live 500 years but then yet again it is light years not regular years i think if you find how man years is 500 light years the that could help you in finding the answer 
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Express the following numbers as decimals: (a) 1.52 x 10^-2, (b) 7.78 x 10^-8, (c) 1 x 10^-6, (d) 1.6001 x 10^3.
Alchen [17]

Answer :

(a) 0.0152

(b) 0.0000000778

(c) 0.000001

(d) 1600.1

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as 5.0\times 10^3

889.9 is written as 8.899\times 10^{-2}

In this examples, 5000 and 889.9 are written in the standard notation and 5.0\times 10^3  and 8.899\times 10^{-2}  are written in the scientific notation.

(a) 1.52\times 10^{-2}

The standard notation is, 0.0152

(b) 7.78\times 10^{-8}

The standard notation is, 0.0000000778

(c) 1\times 10^{-6}

The standard notation is, 0.000001

(d) 1.6001\times 10^{3}

The standard notation is, 1600.1

5 0
4 years ago
2c4H10 +13O2 —>8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10
Elena-2011 [213]

Answer:

\large\boxed{\text{28 mol of O$_{2}$}}

Explanation:

             2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol:      4.3  

13 mol of O₂ react with 2 mol of 2C₄H₁₀

\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}

5 0
3 years ago
Which of the following changes depending on the strength of the gravity field it is in?
jek_recluse [69]
I believe it’s C.) Mass. Hope I’m right.
5 0
3 years ago
What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
Marina86 [1]

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

P = 1.75 atm

n = 2.5 moles

T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

6 0
3 years ago
Help me please
kati45 [8]

Answ????

Explanation:

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5 0
3 years ago
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