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taurus [48]
3 years ago
14

How does heavy water in a nuclear reactor help to keep the chain reaction going?

Chemistry
1 answer:
yarga [219]3 years ago
4 0
<span>Heavy water acts as a neutron moderator. The neutrons formed by fission come off at high velocity, but U-235 absorbs neutrons much more strongly when they are slowed down, so all so-called thermal reactors use a moderator to slow them down until they achieve the average velocity appropriate to the atoms of the moderator which is at approx 300 deg C in a water reactor. Other moderators can be light water or graphite. Heavy water has the advantage over light water that it absorbs fewer neutrons so natural uranium can be used, whilst light water requires enriched uranium fuel. Graphite in a very pure form is also a good moderator but requires a gas coolant, carbon dioxide has been used in magnox and agr reactors, but there is a time limit on the life of the reactor due to graphite corrosion, and these are now obsolete as far as new builds are concerned. The Soviet RBMK reactors used graphite but with a water coolant inside pressure tubes, but after Chernobyl there will certainly be no more of these built
</span>
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14.28 explain how you would distinguish between each pair of compounds using high- resolution mass spectrometry
vagabundo [1.1K]

The distinguish between each pair of compounds using high- resolution mass spectrometry by the exact mass rather than nominal mass are utilizes to measure the compound.

The mass spectrometry is involves the following steps :

  • The ionization
  • acceleration
  • deflection
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Mass spectrometry is the analytical method useful for the calculating the mass to charge ratio ( m / z ). the mass spectrometry is based on the newton's second law and the momentum.

Thus, the mass spectroscopy is method to measure the molecular mass of the compound and indirectly helps examine the isotopes and based on the newton's second law .

To learn more about mass spectroscopy here

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7 0
1 year ago
How many sig figs are there<br> 5.00000008
djverab [1.8K]

Answer:

2 sig figs.

Explanation:

Sig Fig Rules:

Any non-zero digit is a significant figure.

Any zeros between 2 non-zero digits are significant figures.

Trailing zeros after the decimal are significant figures.

6 0
3 years ago
A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec
balu736 [363]

Answer:

E.F= OH

M.F=O_{2} H_{2}

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

\frac{34.00}{17.008}= 1.999 round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- O_{2} H_{2}

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