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3 years ago

11

How can water be used for energy ?

2 answers:

Fantom [35]3 years ago

6 0

First one is B and the second one is C

Sedaia [141]3 years ago

6 0

B for the first question and C for the second question.

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If there is no net force on an object, then the object will

user100 [1]

If there is no net force on an object, then the object will <span>maintain it's rate of speed. Basically, net force is the change in an object's motion. If it is stationary and not moving, the object will stay stationary. If the object is moving at a rate of 2 miles per hour, it will constantly continue to move 2 mph because there is no net force.</span>

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3 years ago

Uranium has an atomic weight of 238 amu.

shepuryov [24]

Answer:

U= 238g/mol

U2O5= 556g/mol

Explanation:

Since U= 238

O=16

U3O5= 2(238)+3(16)=556g/mol

4 0

3 years ago

Please let me know if this is correct?

siniylev [52]

Answer: Yes! you're all good. Alkali metals in group 1 are the most metallic :)

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3 years ago

HOW FAST CAN YOU ANSWER ( new)

GrogVix [38]

Answer:

d

Explanation:

6 0

3 years ago

Read 2 more answers

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago