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wariber [46]
3 years ago
6

I beg you please please I beg you help me please I’m litterly crying please :(

Chemistry
1 answer:
Anna35 [415]3 years ago
4 0
Nitrogen is the answer
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1. Based on the data set and the true measured value being 28km.
motikmotik

Answer:

this data is both accurate and precise

4 0
3 years ago
Read 2 more answers
2NaClO3 ——> 2NaCl + 3O2
Alborosie

Answer:  

<h3>1) 208.09 g</h3><h3>2) 7.11 × 10²² particles</h3>

<h2>Explanation:</h2><h3>NUMBER ONE</h3>

<u>Find moles of Known Substance (NaCl)</u>

Moles of NaCl = mass ÷ molar mass

                        =  114.25 g ÷ 58.44 g/mol

                        = 1.955 mol

<u>Use Mole Ratio of Known to Unknown (NaCl : NaClO₃) to Determine Moles of Unknown</u>

The mole ratio of NaCl to NaClO₃ is 2 to 2 or simply 1 to 1.

This means that for each mole of NaCl produced the same number of moles of NaClO₃ was decomposed.

∴ since moles of NaCl = 1.955 mol,

then moles of NaClO₃ = 1.955 mol

<u>Find Mass of Unknown </u>

Mass of NaClO₃ = moles × molar mass

                           = 1.955 mol × 106.44 g/mol

                           =  208.09 g

<h3>NUMBER TWO</h3>

<u>Determine the Moles of Known (O₂)</u>

moles of O₂ = (volume × density) ÷ molar mass

                    = (68.30 L × 0.0828 g/L) ÷ 32 g/mol

                    = 0.177 mol

<u>Use Mole Ratio of Known to Unknown (O₂ : NaClO₃) to Determine Moles of Unknown</u>

Mole ratio of O₂ : NaClO₃ is 3 : 2. Therefore for each mole of oxygen produced, ²/₃ the moles of NaClO₃ is produced.

∴ if moles of O₂  =  0.177 mol

then moles of NaClO₃ =  0.177 mol × ²/₃

                                     = 0.118 mol

<u>Find the Number of Particles of Unknown</u>

Number of particles  = moles × Avogadro's Number

                                  = 0.118 mol × (6.022 × 10²³ particles/mol)

                                  =  7.11 × 10²² particles

3 0
3 years ago
Determine the mass of 10g of CaCO3​
NISA [10]

Answer:

= 100u. Hence 10 g = 0.1 mole. Hope it's helpful to u

3 0
3 years ago
You’re beautiful *heart*
Georgia [21]
I’m not your just blind. I’m sorry
8 0
3 years ago
A 34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon dioxide and 76.5 grams of
natka813 [3]

Answer:

The answer to your question is  CH₄

Explanation:

Process

1.- Write the possible chemical reaction

                   CxHy + O₂   ⇒    CO₂  +   H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of CO₂ = 12 + 32 = 44 g

                         44 g of CO₂ ------------------- 12 g of carbon

                         93.5 g of CO₂ ----------------  x

                          x = (93.5 x 12) / 44

                          x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

                         18 g of water -------------   2 g of hydrogen

                          76.5 g of water --------    x g of hydrogen

                           x = (76.5 x 2) / 18

                           x = 8.5 g of hydrogen

4.- Calculate the moles of carbon and hydrogen in the sample

                           12g of carbon ------------- 1 mol

                           25.5g of carbon ---------- x

                             x = (25.5 x 1) / 12

                              x = 2.12 moles of carbon

                            1 g of hydrogen ----------- 1 mol

                            8.5 g of hydrogen -------- x

                             x = (8.5 x 1) / 1

                             x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

                                                    CH₄                                        

7 0
3 years ago
Read 2 more answers
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