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ad-work [718]
3 years ago
7

Which set of numbers gives the correct possible values of 1 for n = 3?

Chemistry
1 answer:
Allushta [10]3 years ago
7 0

Answer:

A. 0,1,2

Explanation:

took a test on it and this was the correct answer

You might be interested in
Give me Investigating questions that have dependent variable independent variable and control variable
Masteriza [31]

Answer:

What type of soil filters water best?

Explanation:

dependent variable - the cleanliness of the water after it is filtered

independent variable - soil

control variable - water

(Asking with a <em>please </em>wouldn't hurt -_-)

4 0
3 years ago
A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water, and an ex- cess of NaOH is add
chubhunter [2.5K]

Answer:

The mass percent of aluminum sulfate in the sample is 16.18%.

Explanation:

Mass of the sample = 1.45 g

Al_2SO_3+6NaOH\rightarrow 2Al(OH)_3+3Na_2SO_4

Mass of the precipitate = 0.107 g

Moles of aluminum hydroxide = \frac{0.107 g}{78 g/mol}=0.001372 mol

According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .

Then 0.001372 moles of aluminum hydroxide will be obtained from:

\frac{1}{2}\times 0.001372 mol=0.000686 mol

Mass of 0.000686 moles of aluminum sulfate :

= 0.000686 mol × 342 g/mol = 0.2346 g

The mass percent of aluminum sulfate in the sample:

=\frac{ 0.2346 g}{1.45g}\times 100=16.18\%

5 0
3 years ago
Calculate the mass of 29.8 mL of aluminum, which has a density of 2.00 g/mL.
MaRussiya [10]

Answer:

The answer is

<h2>59.6 g </h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

Density of aluminum = 2.00 g/mL

volume = 29.8 mL

The mass is

mass = 2 × 29.8

We have the final answer as

<h3>59.6 g</h3>

Hope this helps you

5 0
3 years ago
A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr
sergey [27]

Answer:

184.62 ml

Explanation:

Let p_1, v_1, and T_1 be the initial and p_2, v_2, and T_2 be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

p_1=p_2 ...(i)

v_1 = 200 ml

T_1= 26 ^{\circ}C = 273+26 =299 K

T_2= 3 ^{\circ}C = 273+3 =276 K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)

For the final condition,

p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)

Equating equation (i), and (ii)

\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}

\frac{v_1}{T_1}=\frac{v_2}{T_2}  [from equation (i)]

v_2=\frac{T_2}{T_1} \times v_1

Putting all the given values, we have

v_2=\frac{276}{299} \times 200 = 184.62 \; ml

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0
3 years ago
What is the limiting reactant if 12 moles of p4 react with 15 moles of o2?.
brilliants [131]

Answer:

see explanation

Explanation:

To determine limiting reactant divide mole quantities of reactants by the respective coefficient in the balanced equation. The smaller value is the limiting reactant.

    P₄      +     5O₂       => 2P₂O₅

12/1 = 12      15/5 = 3

O₂ is the limiting reactant. P₄ will be in excess when rxn stops.

4 0
2 years ago
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