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andriy [413]
3 years ago
8

a plant fertilizer contains 15% by mass nitrogen. in a container of soluble plant food. there are 10.0 oz of fertilizer . how ma

ny grams of nitrogen are in the container?
Chemistry
1 answer:
trasher [3.6K]3 years ago
7 0
If we convert the ounces to grams, there are approximately 283.495 grams of plant fertiliser

If nitrogen has 15% of this, all we have to do is divide this number by 100 to get the mass of 1% and multiply it by 15.

In the end, we end up with the mass of 42.5243 g

Hope I helped! xx
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Alveoli are tiny sacs of air in the lungs. Their average diameter is 4.50 × 10−5 m. Calculate the uncertainty in the velocity of
Arisa [49]

<u>Answer:</u> The uncertainty in the velocity of oxygen molecule is 4.424\times 10^{-5}m/s

<u>Explanation:</u>

The diameter of the molecule will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

\Delta x.\Delta p=\frac{h}{2\pi}

where,

\Delta x = uncertainty in position = d = 4.50\times 10^{-5}m

\Delta p = uncertainty in momentum  = m\Delta v

m = mass of oxygen molecule = 5.30\times 10^{-26}kg

h = Planck's constant = 6.627\times 10^{-34}kgm^2/s^2

Putting values in above equation, we get:

4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s

Hence, the uncertainty in the velocity of oxygen molecule is 4.424\times 10^{-5}m/s

8 0
3 years ago
Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.​
Elanso [62]

Answer:

Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions =  1.49 × 10²³  +  1.49 × 10²³ = 2.98 × 10²³.

Explanation:

1 mole of any compound contains 6.023 × 10²³ molecules.

molecular weight of NaCl is 23 + 35.5 = 58.5 g.

so, 58.5 grams of NaCl makes 1 mole

⇒ 14.5 g of NaCl = \frac{14.5}{58.5} =  0.248 moles.

⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules

=  1.49 × 10²³ molecules.

And 1 molecule contains 1 Na ion and 1 Cl ion.

⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions =  1.49 × 10²³  +  1.49 × 10²³ = 2.98 × 10²³.

6 0
4 years ago
Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
7 0
3 years ago
A new material formed in a chemical reaction is called a?
fredd [130]
Products


Chemical reactions are characterized by the formation of new products, and the making and breaking of strong chemical bonds.
7 0
3 years ago
A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for
zloy xaker [14]

Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

% abundance of isotope 1 of metal = 72.15% = \frac{72.15}{100}

Mass of isotope 2 of metal= 86.9092

% abundance of isotope 2 of metal = 27.85% = \frac{27.85}{100}

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(84.9118)\times \frac{72.15}{100})+(86.9092)\times \frac{27.85}{100}]]

A=85.47

Therefore, the atomic weight of the metal would be 85.47.

6 0
3 years ago
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