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3 years ago

The diaphragm adjusts the ____________ on a light microscope.

Chemistry

2 answers:

kykrilka [37]3 years ago

5 0

Is used to vary the intensity and size of the cone of light that is projected upward into the slide

AlladinOne [14]3 years ago

5 0

The diaphragm adjusts the amount of light on a light microscope.

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What causes the jet stream? (4 points)

mamaluj [8]

The answer is A.

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kwrob

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3 years ago

How many hydrogen atoms are present in 1.00mg of Aspartame?

Elena-2011 [213]

Answer:

$3.68*10^1^9$ atoms. For every one mole of Aspartame there are 18 moles of hydrogen

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2 years ago

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A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).

katen-ka-za [31]

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

Explanation:

The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:

Vrms = $\sqrt{3RT/M}$

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

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4 years ago

What type of compound is Calcium phosphide?

Maru [420]

Answer:

Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). Calcium phosphide (CP) is the inorganic compound with the formula Ca3P2. It is one of several phosphides of calcium, being described as the salt-like material composed of Ca2+ and P3−.

Explanation:

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3 years ago

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90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$

ΔᵣH° is negative, so the reaction is exothermic.

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3 years ago