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3 years ago

Which of the following is a method for separating a mixture

Chemistry

2 answers:

zvonat [6]3 years ago

8 0

A I’m pretty sure ur welcome

Karo-lina-s [1.5K]3 years ago

5 0

A) Using paper chromatography to separate pigments by density

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A large piece of iron was reacted with an acid. What would happen to the rate of the reaction had the large piece been finely di

Anna007 [38]

The reaction occurs in a similar way as magnesium does, but much less vigorous. Strong heating is required to make iron powder burn in oxygen. The reaction gives out a yellow showery sparks and produces a black solid. iron reacts with dilute hydrocloric acid to give iron chloride and hydrogen gas.

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3 years ago

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Find the [H+] in an acetic acid solution that has a pH of 5.12

Eva8 [605]

Data:
pH = 5.2
[H+] = ?

Knowing that: (Equation to find the pH of a solution)
$pH = -log[H+]$

Solving:
 $pH = -log[H+]$
$5.2 = - log [H+]$
Knowing that the exponential is the opposite operation of the logarithm, then we have:
$[H+] = 10^{-5.2}$
$\boxed{\boxed{[H+] = 6.30*10^{-6}}}\end{array}}\qquad\quad\checkmark$

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3 years ago

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon remain after 7.22 days if the sample initi

Nesterboy [21]

substitute: t1/2=ln(2)k→k=ln(2)t1/2 Into the appropriate equation: [A]t=[A]0∗e−kt [A]t=[A]0∗e−ln(2)t1/2t [A]t=(250.0 g)∗e−ln(2)3.823 days(7.22 days)=67.52 g

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3 years ago

Someone help me , quick

sashaice [31]

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3 years ago

If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti

ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0

3 years ago