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4 years ago

You mix 55 ml of 1.00 m silver nitrate with 25 ml of 1.06 m sodium chloride. what mass of silver chloride should you form?

Chemistry

1 answer:

enot [183]4 years ago

6 0

AgNO3 + NaCl ---> AgCl(s) + NaNO3

find moles of each

mmol = M*V

mmol = 55*1 = 55 mmol of Ag+

mmol = 25 * 1.06 = 26.5 mmol of Cl-

then limitn greactant is Cl

we can produce up to 26.5 mmol of Agcl

mass = mol*MW = (26.5 * 10^-3) * 143.32 = 3.7980 g of AgCl

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Which pair correctly matches the lab equipment with its purpose?

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3 years ago

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An enzyme is discovered that catalyzes the chemical reaction:SAD -------->HAPPY

jekas [21]

Answer: Km = 10μM

Explanation: Michaelis-Menten constant (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: Turnover number (Kcat), which is the number of time an enzyme site converts substrate into product per unit time.

Enzyme veolcity is calculated as:

$V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}$

where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;

To calculate Km:

$V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate]$

$K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}$

$K_{m} = \frac{0.02*600*40-9.6*40}{9.6}$

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The Michaelis-Menten for the substrate SAD is 10μM.

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3 years ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

SSSSS [86.1K]

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

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3 years ago

A mixture of helium and neon gases has a density of 0.2460 g/L at 35.2°C and 288 torr. What is the mole fraction of neon in this

pshichka [43]

We know the volume of one mole of gas at 273 K and 760 Torr is 22.4 L. Using
(PV)/T = constant
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(760 * 22.4) / 273 = (288 V₂) / 308.2
V₂ = 66.7 L
Mass of He: 4
Mass of Ne: 20
Fraction of Ne: x
Fraction of He: 1 - x
avg density = (∑(component fraction × component mass))/volume
0.2460 = (20x + 4(1 - x))/ 66.7
x = 0.775

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4 years ago

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Alja [10]

The best answer is #2

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3 years ago

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