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3 years ago

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Chemistry

1 answer:

natali 33 [55]3 years ago

8 0

Answer:

I would say is B but I do t really know

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Which particle diagram represents a mixture of three substances

Anni [7]

Following the key in the diagram (see the attached image), the only particle diagram that represents a mixture of three substances is diagram 2.

To simplify it, let us replace the key in the diagram as follows;

atom of one element = A
atom of different element = B

Diagram 1 consists of only AA and AB

Diagram 2 consists of AA, BB, and AB.

Diagram 3 consists of AA and ABA

Diagram 4 consists of AA and BAB

Thus, only diagram 2 has a mixture of 3 substances.

More on mixtures can be found here: brainly.com/question/6594631

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3 years ago

if 2.0 g of hydrogen sulfide, H2S(g) reacts with 5.0 g of sodium hydroxide what mass of the excess reactant is present when the

Ronch [10]

Answer: iits 9.g

Explanation:

7 0

3 years ago

How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius

elixir [45]

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

<u>Step 1: </u>Data given

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

<u>Step 2:</u> Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

Q = 50208 Joule = 50.2 kJ

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

8 0

3 years ago

Please help me due today at 7:00pm please help giving more points please help me

kkurt [141]

The answer is 20N as when the object is accelerating at 3 m/s2 it as a force of 10N applied to it so to get to 6m/s2 3(x2) = 6 so 10(x2) = 20

6 0

3 years ago

what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?

MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻ + H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF, the [ OH ⁻] from the given pH, and finally the mass needed to produce that OH⁻ concentration.

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is 6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14 ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

F⁻ HF OH⁻

Equilibrium X 2.51 x 10⁻⁶ 2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X = 1.5 x 10⁻¹¹ ⇒ X = (2.51 x 10⁻⁶)² / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L * 0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF * 42.0 g/mol NaF = 6.03 g NaF

7 0

3 years ago