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professor190 [17]

2 years ago

15

A chef makes a marinade by mixing red pepper flakes, dried onion flakes, and a ground ginger with soy sauce. He sets the marinad

e on the counter for a few minutes. When he returns, he must stir the marinade because the spices have settled to the bottom of the container.
answer: the mixture is a suspension

2 answers:

Sophie [7]2 years ago

6 0

Answer:

Explanation: suspension is the answer

Alisiya [41]2 years ago

3 0

Answer:

Suspension

Explanation:

This mixture is a simple suspension.

A suspension is a mixture of small insoluble particles of a solid in a liquid or gas. Here, it is insoluble particles in liquid.

Suspensions are settle on standing this is why they have to be mixed again.
The particles do not pass through ordinary filter paper.
They are usually cloudy and have an opaque color.
The marinade is simply a suspension.
It is not a solution because they do not settle on standing.
Also, colloids do not settle on standing.

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Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,

vfiekz [6]

Answer:

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

8 0

1 year ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

2 years ago

In the figure shown, AD, CE, and FB intersect at point F.

EleoNora [17]

Answer:

you need to send us the figure

Explanation:

7 0

1 year ago

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

5 0

1 year ago

A 500.0 ml buffer solution is 0.10 m in benzoic acid and 0.10 m in sodium benzoate and has an initial ph of 4.19. part a what is

slavikrds [6]

Initial moles of C₆H₅COOH = 500/1000 × 0.10 = 0.05mol
Initial moles of C₆H₅COONa = 500/1000 × 0.10 = 0.05 mol
initial pH = Pka + log([C₆H₅COONa/ moles of C₆H₅COOH)
4.19 = pKa + log(0.05/0.05)
→pKa = 4.19
C₆H₅COOH + NaOH → C₆H₅COONa ₊ H₂o
moles of NaOH added = 0.010 mol
moles of C₆H₅COOH = 0.05 - 0.025 = 0.025 mol
Final pH = pKa + log([C₆H₅COONa)/[ C₆H₅COOH])
=pKa + log(moles of C₆H₅COONa/moles of C₆H₅COOH)
= 4.19 + log(0.025/0.075)
4.29

8 0

1 year ago