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2 years ago

A single gram of 235U fuel can heat and boil 2.49 x 107 g of water. What volume of water is this?

1 answer:

iogann1982 [59]2 years ago

6 0

You just have to convert the mass of water into volume.

To do that you use the density of water, which is about 1.0 g/ ml

So, from the formula of density D = M / V, you get V = M / D

=> V = 2.49 * 10^7 grams / 1.0 g / ml = 2.49 * 10 ^ 7 ml

You can pass that to liters using the conversion factor 1000 ml = 1 l

2.49 * 10^7 ml * 1 l / 1000 ml = 2.49 * 10^4 l = 24,900 l

Answer: 24,900 l

You might be interested in

The number of oxygen atoms in 48.0g of oxygen gas

Doss [256]

Answer:

Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.

After that if we to multiply the avogadro number with it. So 3 *NA

Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.

So final claculation is =3*2*NA.

Explanation:

your welcome

brainlest PLEASEEEEEEE!

3 0

3 years ago

The atomic number of Re is 75. The atomic mass of one of its isotopes is 186. How many neutrons are in an atom of this isotope?

USPshnik [31]

Answer:

B.111

Explanation:

Atomic mass = number of protons(or atomic number) + number of neutrons

186 = 75 + number of neutrons

number of neutrons = 186 -75 = 111

7 0

2 years ago

Through an uplift under the earth crust at a dirvergent boundary a _____ is formed.

postnew [5]

Out of the following given choices;

a. cliff b. fault

c. plateau d. mountain

The answer is b. A divergent boundary is a line at which two tectonic plates are moving away from each other. It is caused by the two magma convection currents in the mantle moving in opposing directions (one clockwise, the other anti-clockwise) hence dragging the crust with them. Therefore the biggest force at the boundary on the crust is that of pulling. This causes fractures and faults on the earth’s crust.

6 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.