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Slav-nsk [51]
2 years ago
6

Are bath bombs pure substances or mixtures

Chemistry
2 answers:
mezya [45]2 years ago
3 0

Answer:

Yes.

Explanation:

Bath bombs are usually made from three key ingredients: baking soda, citric acid and cornstarch, said Frankie Wood-Black, an instructor in chemistry at Northern Oklahoma College and an experienced bath-bomb maker. Often, bath bombs also include dyes and perfumes, and sometimes they have epsom salt.

evablogger [386]2 years ago
3 0

Answer: yes

Explanation:

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Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc
jeka94

Answer:

T_f = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

T_i = 21.5°C

C_{calorimeter} = 6.15 kJ/°C

T_f = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate ×  \frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

=  0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter =  C_{calorimeter} (T_f-T_i} )

21.84 kJ =6.15 kJ/°C * (T_f-21,5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - 132.225 kJ

21.84 KJ + 132.225 kJ = (6.15 kJ/^0C * T_f)

154.065 kJ = (6.15 kJ/^0C * T_f)

T_f = \frac{154.065kJ}{6.15kJ/^0C}

T_f =25.05°C

4 0
3 years ago
The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea
sineoko [7]
We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula: 
Kc   = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 =  0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0
3 years ago
The process though which animals and plants produce offspring
ki77a [65]

Answer:

reproduction

Explanation:

can be asexual or sexual

7 0
2 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
Metals give up electrons to become cation or anion
galina1969 [7]
Metals usually become cations since electrons are negatively charged and when they are lost there are more protons than electrons making a positive net charge in the atom.  (cations are positively charged ions while anions are negatively charged ions)

I hope this helps.  Let me know if anything is unclear.
8 0
3 years ago
Read 2 more answers
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