Mass of solute = 35 grams
Mass of solution = mass of nacl + mass of water = 35.8 + 184 = 219.8 grams

Concentration = mass of Nacl x 100/mass of solution.

= 35x100/219.8 = 15.92%.

Hope this helps you.

7 0

4 years ago

Given two temperatures, 36°F and 72°F, which of the following correctly compares the number of opportunities for particles to al

Ivanshal [37]

The answer is C) greater at 72 °F

hope it helped

3 0

3 years ago

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Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the

Norma-Jean [14]

Answer:

9.18g

Explanation:

Step 1: Write the reduction half-reaction

Au³⁺(aq) + 3 e⁻ ⇒ Au(s)

Step 2: Calculate the mass of gold is produced when 15.0A of current are passed through a gold solution for 15.0min

We will use the following relationships:

1 min = 60 s

1 A = 1 C/s

1 mole of electrons has a charge of 96486 C (Faraday's constant).

1 mole of Au is produced when 3 moles of electrons circulate.

The molar mass of Au is 196.97 g/mol.

The mass of gold produced is:

$15.0 min \times \frac{60s}{1 min} \times \frac{15.0C}{1s} \times \frac{1 mol e^{-} }{96486C} \times \frac{1molAu}{3 mol e^{-} } \times \frac{196.97gAu}{1molAu} = 9.18gAu$

3 0

3 years ago

What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol)?

NemiM [27]

the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g

The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn

3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2

the formula is n= mass/M so, now substituting values

m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3

so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g

so mass of aluminum oxide obtained = 1.36g

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3 0

2 years ago