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3 years ago

Please help I will give ALOT OF BRAINLY POINTS!!!!!

Chemistry

2 answers:

d1i1m1o1n [39]3 years ago

7 0

The answer to the question is A

solmaris [256]3 years ago

6 0

I believe the answer is mitosis

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Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.

Thepotemich [5.8K]

Answer:

m = 2.955x10⁻² mol/kg

X = 5.323x10⁻⁴ mol NaCl/Total moles

(w/w)% = 0.1726%

ppm = 1726 mg/kg

Explanation:

Molality is the ratio between moles of solute per kg of solution.

As the solution is 2.950×10⁻² mol/L, mililters are 999.2mL and density is 0.9982 g/mL, molality is:

m = 2.950×10⁻² mol/L×(1L/0.9982kg) = 2.955x10⁻² mol/kg

Mole fraction is moles of NaCl/total moles.

Moles of H₂O are:

999.2mL×(0.9982g/mL)×(1mol/18,02g) = 55,35 moles of H₂O

Moles of NaCl are:

2.950×10⁻² mol/L×(0.9992L)= 2.950×10⁻² mol of NaCl

mole fraction is:

X = 2.950×10⁻² mol of NaCl / (2.950×10⁻² mol of NaCl+55.35mol water) = 5.323x10⁻⁴ mol NaCl/Total moles

Mass of NaCl is:

2.950×10⁻² mol of NaCl×(58.44g/mol) = 1.724g of NaCl

Mass of water is:

55.35mol water×(18.02g/mol) = 997.4g of H₂O

(w/w)% is:

1.724g of NaCl / (1.724g of NaCl+997.4g of H₂O) ×100 = 0,1726%

Parts per million is mg of NaCl per kg of solution, that is:

1724mg of NaCl / 0.999124g = 1726 ppm

I hope it helps!

5 0

2 years ago

2-ethyl-4,4-dimethyl-2-octanone. Draw the structure

Dvinal [7]

Is there a picture maybe? I can’t help you

5 0

2 years ago

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

Mandarinka [93]

Answer:

$\boxed{\text{-55.8 kJ/mol NaOH}}$

Explanation:

NaOH + HNO₃ ⟶ NaNO₃ + H₂O

There are two energy flows in this reaction.

$\begin{array}{cccl}\text{Heat from neutralization} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\DeltaH & + & mC\Delta T & =0\\\end{array}$

Data:

V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹

V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹

T₁ = 35.00 °C; T₂ = 37.00 °C

Calculations:

(a) q₁

$n_{\text{NaOH}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}\\\\n_{\text{HNO}_{3}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}$

We have equimolar amounts of NaOH and HNO₃

n = 0.0300 mol

q₁ = 0.0300ΔH

(b) q₂

V = 100.0 mL + 100.0 mL = 200.0 mL

m = 200.0 g

ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C

q₂ = 200.0 × 4.184 × 2.00 = 1674 J

0.0300ΔH + 1674 = 0

0.0300ΔH = -1674

ΔH = -1674/0.0300

ΔH = -55 800 J/mol

ΔH = -55.8 kJ/mol

$\Delta_{r}H^{\circ} = \boxed{\textbf{-55.8 kJ/mol NaOH}}$

6 0

3 years ago

This is really confusing

Sliva [168]

I’m like 85% sure that’s correct

4 0

3 years ago

At a job interview, the most effective way to prove your proficiency to a prospective employer is to show that you have

inna [77]

The most effective way to showcase your proficiency in a job interview is to show certificates. For example, in this case, option D. certification as an electronics technician. Option A's online research experience is menial as well as using own test equipment as this renders hassle and futile. Membership cannot guarantee job proficiency, too. Answer is D.

7 0

3 years ago