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3 years ago

Which of these equations best summarizes photosynthesis?

Chemistry

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer: C

Explanation:

Sunlight

6 CO2 + 6 H2O --------------> C6H12O6 + 6 O2

Chlorophyll

Carbon(IV)oxide Water Glucose Oxygen

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Rust is a compound with the formula F e 2 O 3 . Which elements combine to form rust?

DanielleElmas [232]

Remembering the periodic table, Fe is iron and O is oxygen. The elements that combine to create rust are iron and oxygen, with the chemical compound being (Iron)2(Oxygen)3.

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3 years ago

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in this cut-away image of an atom, where is the red arrow most likely pointed at the greatest probability of locating an electro

emmasim [6.3K]

Answer:

Explanation:

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3 years ago

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How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$

For neutralization, equal number of moles of $H^+$ ions will neutralize same number of $OH^-$ ions.

$\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles$

As, $H_2SO_4\rightarrow 2H^++SO^{2-}_4$

From this reaction, we conclude that

2 moles of $H^+$ ion is given by the 1 mole of $H_2SO_4$

$3309.8\times 10^{-9}$ moles of $H^+$ ion is given by $\frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9}$ moles of $H_2SO_4$

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of $H_2SO_4$ × Molar mass of sulfuric acid

Mass of sulfuric acid = $(1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g$

Therefore, the mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

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3 years ago

Zirconium has an atomic number of 40, And an atomic mass of 91. How many electrons does zirconium have?

Vedmedyk [2.9K]

Answer: 40

Explanation: Whatever the atomic number is, that’s also the number of electrons and protons.

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3 years ago

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Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago