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3 years ago

Which layers of the earths are common to both the compostitional and mechanical descriptions

Chemistry

1 answer:

Nutka1998 [239]3 years ago

3 0

The correct is answer is c

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2. How many nanoliters are in 2.87 x 10-10 gallons?

Morgarella [4.7K]

Answer:

$1.09nL$

Explanation:

Hello,

In this case, given that 1 gal equals 4 qt, 1 qt equals 0.9464 L and 1 L equals 1x10⁹ nL, the dimensional analysis turns out:

$2.87x10^{-10}gal*\frac{4qt}{1gal} *\frac{0.9464L}{1qt}*\frac{1x10^9nL}{1L}\\ \\1.09nL$

Best regards.

4 0

3 years ago

Which bond is the weakest????<br> Please hurry!!,<br> I’ll put a pic.

sammy [17]

B because it’s got the least connections

5 0

2 years ago

Read 2 more answers

A screw is a type of simple machine. If we look closely at a screw, we see that it is really a _________ wrapped around a centra

krek1111 [17]

I believe the answer in an inclined plane, because the ridge running along the sides run up the core. (Also, the other two answers are incorrect, so process of elimination)

5 0

3 years ago

Read 2 more answers

Consider the balanced equation below.

Bumek [7]

Answer : The correct option is, (2) 8 mole of hydrogen will react with 1 mole of sulfur.

Explanation :

The balanced chemical equation is,

$8H_2+S_8\rightarrow 8H_2S$

By the stoichiometry we conclude that, 8 moles of hydrogen react with 1 mole of sulfur to give 8 moles of hydrogen sulfide.

Hence, the correct options is, (2) 8 mole of hydrogen will react with 1 mole of sulfur.

6 0

3 years ago

Read 2 more answers

onsider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a c

Juli2301 [7.4K]

Answer:

Explanation:

From the given information;

Let Q(t) = mass of dye in the tank as a function of time

The mass in the tank = 200 L × (1g/L) = 200 g

Using the law of mass conservation;

$\dfrac{dQ}{dt} = \text{(Rate of incoming mass)- (rate of outgoing mass)}$

$\dfrac{dQ}{dt} = (2 L/min ) (0 \ g/L) - (2 L/min ) (\dfrac{Q}{200}g/L)$

$Q' = \dfrac{-Q}{1000}$

Q(0) = 200

By finding the solution to the ODE using the method of separation of variables;

$\dfrac{Q'}{Q} = -0.01$

$Q(t) = Ce^{-0.01t}$

Using the initial condition;

200 = Q(0) = C

$Q(t) = 200e^{-0.01t}$

1% of 200g = 2g of dye solution

∴

$2 = 200e^{-0.01t}$

$e^{-0.01t}=0.01$

$t =\dfrac{ In(0.01}{-0.01}$

t = 460.5 hours

4 0

3 years ago