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4 years ago

6

A chemist needs 40.0 g of concentrated sulfuric acid for an experiment. the density of concentrated sulfuric acid at room temper

ature is 1.84 g/ml. what volume of the acid is required?

1 answer:

STALIN [3.7K]4 years ago

4 0

Hope this helps you;):)

You might be interested in

H e l p,,, ill give brainliest :)

mote1985 [20]

Hi there! :)

Both are compounds.

$CO_{2}$ contains 1 carbon (C) atom and 2 oxygen (O) atoms.

It contains two different elements, so it is a COMPOUND.

$Na_{2} O$ is an IONIC compound with elements Sodium (Na) and Oxygen (O), which are different elements. Therefore, it is also a COMPOUND.

6 0

3 years ago

Which of these statements describes this mathematical equation?

salantis [7]

Answer:

The final pressure of a gas is inversely proportional to the volume change and directly proportional to temperature

Explanation:

Given

$\frac{P_1V_1}{T_1} = k$

Required

Interpret

$\frac{P_1V_1}{T_1} = k$

Multiply both sides by T1

$T_1 * \frac{P_1V_1}{T_1} = k * T_1$

$P_1V_1 = kT_1$

Divide both sides by V1

$\frac{P_1V_1}{V_1} = \frac{kT_1}{V_1}$

$P_1 = \frac{kT_1}{V_1}$

This can be rewritten as:

$P_1 = k\frac{T_1}{V_1}$

In the above expression; k is a constant of proportionality.

So, the equation can be written as variation as follows:

$P_1\ \alpha\ \ \frac{T_1}{V_1}$

To interpret:

<em>P varies directly to T (the numerator) and inversely to V (the denominator).</em>

<em></em>

3 0

3 years ago

A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t

Serga [27]

Explanation:

1)

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

$1.00 M=\frac{m}{40 g/mol\times 1.00 L}$

$m=1.00 M\times 40 g/mol\times 1.00 L = 40. g$

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

2)

$M_1V_1=M_2V_2$ (dilution equation)

Molarity of the NaOH solution = $M_1=2.00 M$

Volume of the solution = $V_1=?$

Molarity of the NaOH solution after dilution = $M_2=1.00 M$

Volume of NaOH solution after dilution= $V_2=1 L$

$M_1V_1=M_2V_2$

$V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L$

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

8 0

3 years ago

garik1379 [7]

Explanation:

1.

Given parameters:

Frequency of the radiation = 8.4 x 10¹⁴Hz

Unknown:

Energy of the wave = ?

Solution:

The energy of a wave is given by the expression below;

E = hf

E is the energy

h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s

f is the frequency

Now insert the parameters and solve;

E = 6.63 x 10⁻³⁴m²kg/s x 8.4 x 10¹⁴Hz

E = 5.57 x 10¹ x 10⁻²⁰J

E = 5.57 x 10⁻¹⁹J

2.

Given parameters:

Wavelength = 2.13 x 10⁻¹³m

Unknown:

Frequency of the wave = ?

Solution:

The frequency of a wave can be determined using the expression;

C = f∧

C is the speed of light = 3 x 10⁸m/s

f is the frequency

∧ is the wavelength

f = $\frac{C}{wavelength}$ = $\frac{3 x 10^{8} }{2.13 x 10^{-13} }$ = 1.41 x 10²¹hz

5 0

3 years ago

During a laboratory experiment, 36.12 grams of Aboz was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What

kvv77 [185]

Answer:

8.70 liters

Explanation:

3O₂+ 4Al → 2AI₂O₃

First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:

36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃

Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:

0.354 mol AI₂O₃ * $\frac{3molO_2}{2molAl_2O_3}$ = 0.531 mol O₂

We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:

1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K

V = 8.708 L

5 0

3 years ago