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Jobisdone [24]
2 years ago
7

WOULD YOU TRUST ME WITH YOUR NUMB ER

Chemistry
2 answers:
vlabodo [156]2 years ago
3 0

Answer:

NOOOOO

Explanation:

BECAUSE NOOOOOOOOO

laila [671]2 years ago
3 0

Answer:

idk

Explanation:

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The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.
Naddika [18.5K]

Answer:

12.6.

Explanation:

  • We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

  • It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

4 0
3 years ago
When combined with alcohol some over the counter drugs can:
Marianna [84]

Can cause a chemacal reaction

can i have brainliest i need it

7 0
3 years ago
Read 2 more answers
Which of the following pairs of elements is most likely to form an ionic compound? a oxygen and fluorine b sodium and aluminum c
irina [24]

An ionic compound consists of a metal AND a non-metal.

<em>Option A:</em>

Oxygen and fluorine are non-metals.

<em>Option B:</em>

Sodium and aluminium are non-metals.

<em>Option C:</em>

Calcium is a metal and chlorine is a non-metal.

<em>Option D:</em>

Nitrogen and sulfur are non-metals.

Thus, the answer is C.

4 0
3 years ago
12. In the modern periodic table, which of the following describes the elements with similar
V125BC [204]
B. same group I believe
8 0
3 years ago
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A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the
likoan [24]

Answer:

The final temperature of the solution is 44.8 °C

Explanation:

assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:

Q dis + Q sol = 0

Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C)  is q dis= -83.3 KJ/mol . And the molecular weight is

M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol

Q dis = q dis * n  = q dis * m/M =  -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ

Qdis= -9.84 KJ

Also Qsol = ms * Cs * (T - Ti)

therefore

ms * Cs * (T - Ti) + Qdis = 0

T=  Ti - Qdis * (ms * Cs )^-1   =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ

T= 44.8 °C

7 0
3 years ago
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