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2 years ago

Convert 3.8 Km/sec to miles/year

1 answer:

5 0

We are asked to convert from units of kilometer per second to units of miles per year. To do this, we need a conversion factor which would relate the different units involved. We either multiply or divide this certain value to the original measurement depending on what is asked. From literature, we will find that 1 mile is equal to 1609 meters and 1000 m is equal to 1 kilometer. Also, we will find that 3600 s is equal to 1 hr, 24 hr is equal to 1 day and 365 days is equal to 1 year. We do the conversion as follows:

3.8 km / s ( 1000 m / 1 km ) ( 1 mile / 1609 meters ) ( 3600 s / 1 hr ) ( 24 hr / 1 day ) ( 365 days / 1 year ) = 74479055.3 miles per year

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You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point

Hoochie [10]

First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

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2 years ago

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A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were r

Ivan

Mass of water released =

$92.8 g CuSO_{4}.5H_{2}O$ × $\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}$

= 33.45 g $H_{2}O$

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2 years ago

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Jill is doing an experiment on the movement of pill bugs. She will place the pill bugs on flat surfaces covered with diffirent m

Ostrovityanka [42]

Answer:D

Explanation:

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2 years ago

How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?

Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

Moles of $NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles$

$W_s$ = weight of the solvent in g = ?

$0.452=\frac{0.348\times 1000}{W_s}$

$W_s=770g$

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

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2 years ago

The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d

OLga [1]

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n = 8.04*10^-4 mol/L *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

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2 years ago