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3 years ago

R-COOH compounds behave as _____. A) bases B) acids C) esters D) ketones

Chemistry

1 answer:

lakkis [162]3 years ago

6 0

<span>R-COOH
-COOH is the group, that shows that the substance is an organic acid.

Answer is B. Acids</span>

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Really need help with this! Chemistry

Westkost [7]

Answer:

a) 0,5

Explanation:

If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5

5 0

2 years ago

Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume wou

lesya [120]

First, let's compute the number of moles in the system assuming ideal gas behavior.

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles

At standard conditions, the standard molar volume is 22.4 L/mol. Thus,

Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>

3 0

4 years ago

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

5 0

3 years ago

2. In the reaction NO + NO2 ⇌ N2O3, an experiment finds equilibrium concentrations of [NO] = 3.8 M, [NO2] = 3.9 M, and [N2O3] =

notsponge [240]

Kc= concentration of product divided by concentration of reactant
NO + NO2 ----> N2O3

Kc =(N2O3) / (No)(NO2)

Kc= ( 1.3 )/{ (3.9)(3.8) }

Kc=0.088 ( answer B)

4 0

3 years ago

Read 2 more answers

A reaction produces 10.5 L if oxygen, but was supposed to produce 1 mol of oxygen. What is the percent yield ? (One mole of any

Rzqust [24]

To answer this item, we assume that oxygen behaves ideally such that it is able to fulfill the following equation,

PV = nRT

If we are to retain constant the variable n and V.

The percent yield can therefore be solved through the following calculation,

n = (10.5 L)/(22.4 L) x 100%

Simplifying,
n = 46.875%

Answer: 48.87%

5 0

3 years ago