The main process contributing material in solution in sediment load is

vredina [299]

Answer:

Electron Configuration and Oxidation States of Neon. Electron configuration of Neon is [He] 2s2 2p6. Possible oxidation states are 0. so the answer is 0.

Explanation:

7 0

4 years ago

How many grams of hydrogen are in 92 g of ch4o?

IRISSAK [1]

The answer to this question would be: 11.5 grams

The molecular mass of hydrogen is 1, oxygen is 16 and carbon is 12. Then the molecular mass of CH4O would be: 12 + 4(1) + 16= 32.
To find the hydrogen mass, you just need to divide the hydrogen with total mass, resulting in the mass ratio of hydrogen to the compound. Then the amount of hydrogen inside 92 gram of CH4O are: 92grams * (4/32)= 11.5 grams

7 0

3 years ago

Read 2 more answers

Brainliest please help

Yuri [45]

1 3/4 Ounces or 1.75 Ounces!

Basically 21/12 = 1.75!

(Mark me as the brainliest if this helps!)

3 0

3 years ago

What is the balanced equation for the combustion of sulfur

Alex17521 [72]

Answer:

There are three possible chemical equations for the combustion of sulfur:

2S (s) + O₂ (g) → 2SO (g)

S (s) + O₂ (g) → SO₂ (g)

2S (s) + 3O₂ (g) → 2SO₃ (g)

Explanation:

Combustion is a reaction with oxygen. The products of the reaction are oxides, and energy is released in the form of heat and light.

Sulfur iis a nonmetal, so the oxide formed is a nonmetal oxide.

The most common oxidation numbers of sulfur are -2, + 2, + 4, and + 6.

The combination of sulfur with oxygen may be only with the positive oxidation numbers (+2, + 4, and +6).

Then you have three different equations for sulfur combustion:

1) Oxidation number +2:

S(s) + O₂(g) → SO(g)

Which when balanced is: 2S(g) + O₂(g) → 2SO(g)

2) Oxitation number +4:

S(s) + O₂(g) → SO₂(g)

That equation is already balanced.

3) Oxidation number +6:

S(s) + O₂(g) → SO₃(g)

Which when balanced is: 2S(s) + 3O₂(g) → 2SO₃(g)

3 0

4 years ago

The common titanium alloy known as T-64 has a composition of 90 weight% titanium 6 wt% aluminum and 4 wt% vanadium. Calculate th

Anna007 [38]

Explanation:

Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.

Mass of titanium = 90 g

Moles of titanium = $\frac{90 g}{47.87 g/mol}=1.8800 mol$

Total number of atoms of titanium , $a_t=1.8800 mol\times N_A$

Mass of aluminum = 6 g

Moles of aluminium = $\frac{6 g}{26.98 g/mol}=0.2223 mol$

Total number of atoms of aluminium, $a_a=0.2223 mol\times N_A$

Mass of vanadium = 4 g

Moles of vanadium= $\frac{4 g}{50.94 g/mol}=0.0785 mol$

Total number of atoms of vanadium $a_v=0.0785 mol\times N_A$

Total number of atoms in an alloy = $a_t+a_a+a_v$

Atomic percentage:

$Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100$

Atomic percentage of titanium:

: $\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%$

Atomic percentage of Aluminium:

: $\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%$

Atomic percentage of vanadium

: $\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%$

6 0

4 years ago