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Katyanochek1 [597]
3 years ago
6

I will give branliest!!!!!! PLEASE ANSWER SOOON! ANSWER ALL QUESTIONS PLEASE EXTRA POINTS TOO

Chemistry
2 answers:
Luda [366]3 years ago
6 0
4.c,5.d.6a good luck
just olya [345]3 years ago
4 0

Answer:

#4 is C

#5 is D

#6 is A

bruh this is easy

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Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai
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Answer:

Explanation:

check the attachment below

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2 years ago
Vanillin is the compound containing carbon, hydrogen, and oxygen that gives vanilla beans their distinctive flavor. The combusti
nignag [31]
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C 14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O molar mass of C=12 g/mole molar mass of H=1 g/mole 0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C 0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O molar mass of O=16 g/mole 9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602 C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602 C2.66H2.66O1 is the empirical formula; to obtain whole numbers multiply by 3 3[C2.66H2.66O1] = C8H8O3 above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu The empirical formula weight and the molecular formula weight are the same . Molecular formula is C8H8O3.</span>
7 0
3 years ago
How many grams are 3.01 × 1023 molecules of CuSO4?
zvonat [6]
Answer is: 79.8 grams of copper(II) sulfate.
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n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
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4 0
3 years ago
A solute is a strong electrolyte that ionizes into 2 ions. What is the molar mass of this solute if 30.76 grams added to exactly
Whitepunk [10]

Answer:

34.9 g/mol is the molar mass for this solute

Explanation:

Formula for boiling point elevation: ΔT = Kb . m . i

ΔT = Temperatures 's difference between pure solvent and solution → 0.899°C

Kb = Ebullioscopic constant → 0.511°C/m

m = molality (moles of solute/1kg of solvent)

i = 2 → The solute is a strong electrolyte that ionizes into 2 ions

For example: AB ⇒ A⁺  +  B⁻

Let's replace → 0.899°C = 0.511 °C/m . m . 2

0.899°C / 0.511 m/°C . 2 = m → 0.879 molal

This moles corresponds to 1 kg of solvent. Let's determine the molar mass

Molar mass (g/mol) → 30.76 g / 0.879 mol = 34.9 g/mol

4 0
3 years ago
A sample of gas occupies a volume of 445 mL at a temperature of 274 K. At what temperature will this sample of gas occupy a volu
Ksenya-84 [330]
V1 = 445ml V2 = 499ml
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By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K
4 0
3 years ago
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