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Flura [38]
3 years ago
12

How many compounds can I make with sodium, hydrogen, oxygen?

Chemistry
2 answers:
yanalaym [24]3 years ago
7 0

Answer:

2H2 + O2 → 2H2O

Explanation:

Natalka [10]3 years ago
6 0

Answer:

маны

Explanation:

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Which one is a decomposition reaction?​
vodomira [7]

Answer:

b no

Explanation:

because it is decomposing into two elements

8 0
3 years ago
Read 2 more answers
I need the answers? Pls help!
Vesnalui [34]
Im pretty sure its h2o so like the 2nd one
4 0
3 years ago
A 5.000 g sample of Niso4 H2O decomposed to give 2.755 g of anhydrous NiSO4.
Vinvika [58]

Answer:

a) 7.0.

b) Nickel sulfate hepta hydrate.

c) 280.83 g/mol.

d) 44.9%.

Explanation:

<u><em>a) What is the formula of the hydrate?</em></u>

The mass of the hydrated sample (NiSO₄.xH₂O) = 5.0 g,

The mass of the anhydrous salt (NiSO₄) = 2.755 g,

The mass of water = 5.0 g - 2.755 g = 2.245 g.

∴ no. of moles of water = mass/molar mass = (2.245 g)/(18.0 g/mol) = 0.1247 mol.

∴ no. of moles of anhydrous salt (NiSO₄) = mass/molar mass = (2.755 g)/(154.75 g/mol) = 0.0178 mol.

∴ water of crystallization in the sample (x) = no. of moles of water/no. of moles of anhydrous salt (NiSO₄) = (0.1247 mol)/(0.0178 mol) = 7.0.

<u><em>b) What is the full chemical name for the hydrate?</em></u>

The name of the salt (NiSO₄.7H₂O) is Nickel sulfate hepta hydrate.

<u><em>c) What is the molar mass of the hydrate? </em></u>

(NiSO₄.7H₂O)

The molar mass = molar mass of NiSO₄ + 7(molar mass of H₂O) = (154.75 g/mol) + 7(18.0 g/mol) = 280.83 g/mol.

<em><u>d) What is the mass % of water in the hydrate?</u></em>

The mass % of water = (mass of water)/(mass of hydrated sample) x 100 = (2.245 g)/(5.0 g) x 100 = 44.9%.

8 0
4 years ago
"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the
Aleksandr-060686 [28]

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = 4 \pi r^{2}

                        = 4 \times 3.14 \times (0.50 m)^{2}

                        = 3.14 m^{2}

    T_{a} = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = esA(T^{4} - T^{4}_{a})

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = 5.6703 \times 10^{-8} Wm^{-2} K^{-4}

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = esA(T^{4} - T^{4}_{a})

                     = 0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

8 0
3 years ago
Which of these are a body fossil A.dinosaur nest B.shrimp burrow C.leaf imprint D.claw print
Darya [45]

The answer is C. leaf imprint


8 0
4 years ago
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