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Svetach [21]
3 years ago
14

Help! 7th Grade Math!

Chemistry
1 answer:
vagabundo [1.1K]3 years ago
6 0

Answer:

D

Explanation: Trust me

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Multiply the atomic weight
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  • A)Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)

  • 2KBr(aq)+BaI2(aq) → 2KI(aq)+BaBr2(s)

  • B)Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid: 
  • CaCO3+ HCl -> CaCl2 + CO2 + H2O To balance chemical equations we need to look at each element individually on both sides of the equation. calcium carbonate is a chemical compound with the formula CaCO3.

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3 years ago
"Calculate the number of millimoles in 500 mg of each of the following amino acids: alanine (MW = 89), leucine (131), tryptophan
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Answer:

- Alanine =  5.61 mmoles

- Leucine = 3.81 mmoles

- Tryptophan = 2.45 mmoles

- Cysteine = 4.13 mmoles

- Glutamic acid = 3.40 mmoles

Explanation:

Mass / Molar mass = Moles

Milimoles = Mol . 1000

500 mg / 1000 = 0.5 g

- Alanine = 0.5 g / 89 g/m → 5.61x10⁻³ moles  . 1000 = 5.61mmoles

- Leucine = 0.5 g / 131 g/m → 3.81 x10⁻³ moles  . 1000 = 3.81 mmoles

- Tryptophan = 0.5 g / 204 g/m →  2.45x10⁻³ moles . 1000 = 2.45 mmoles

- Cysteine = 0.5 g / 121 g/m → 4.13x10⁻³ moles . 1000 = 4.13 mmoles

- Glutamic acid = 0.5 g 147 g/m → 3.40x10⁻³ moles . 1000 =  3.4 mmoles

5 0
3 years ago
How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
SpyIntel [72]

Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as 5.11 \times 10^{-5} \, s^{-1}

So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

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<h3><u>Answer;</u></h3>

exceeds evaporation over land

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