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SVETLANKA909090 [29]
2 years ago
5

What is the molarity of a stock solution if 60 mL were used to make 150 mlof a .5M solution? ​

Chemistry
1 answer:
olga2289 [7]2 years ago
4 0

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $\frac{V1M1}{V2}

Plugin the values as,

M2 = $\frac{150 \times 0.5}{60}

       = 1.25 M

So the molarity of the stock solution is 1.25 M.

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Consider the following types of electromagnetic radiation:
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Explanation:

Electromagnetic wave              Wavelength

(1) Microwave  =   1 m to 1 mm = 10^9 nm to 10^6 nm

(2) Ultraviolet  =    10 nm to 400 nm

(3) Radio waves  =   1 mm to 100 km = 10^6 nm to 10^{14}nm

(4) Infrared  =    700 nm to 1 mm

(5) X-ray  =   0.01 nm to 10 nm

(6) Visible =   400 nm t0 700 nm

a) In order of increasing wavelength:

: 5 < 2 < 6 < 4 < 1 < 3

b) Frequency of the electromagnetic wave given as:

\nu=\frac{c}{\lambda }

\nu = frequency

\lambda = Wavelength

c = speed of light

\nu \propto \frac{1}{\lambda }

So, the increasing order of frequency:

: 3 < 1 < 4 < 6 < 2 < 5

c) Energy(E) of the electromagnetic wave is given by Planck's equation :

E=h\nu

E\propto \nu

So, the increasing order of energy:

: 3 < 1 < 4 < 6 < 2 < 5

6 0
3 years ago
Calculate the volume occupied by 272g
cupoosta [38]

The answer for the following problem is mentioned below.

<u><em>Therefore volume occupied by methane gas is  184.78 × 10^-3 liters </em></u>

Explanation:

Given:

mass of methane(CH_{4}) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of  methane(CH_{4}) = 16.0​  grams

We know;

According to the ideal gas equation,

<u><em>P × V = n × R × T</em></u>

here,

n = m÷M

n =272 ÷ 16

<u><em>n = 17 moles</em></u>

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

<u><em>Therefore volume occupied by methane gas is  184.78 × 10^-3 liters </em></u>

<u><em></em></u>

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6 0
2 years ago
While you were "sweating" your chemistry test, water vapor evaporates from your body, absorbing 50,000 J of energy. (assume no
TiliK225 [7]

Answer:

\boxed {\boxed {\sf B. \ 22 \ grams}}

Explanation:

We need to use the formula for heat of vaporization.

Q=H_{vap}*m

Identify the variables.

  • The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
  • Q is the energy, in this problem, 50,000 Joules.
  • m is the mass, which is unknown.

H_{vap}=2260 \ J/g\\Q=50,000 \ J \\

Substitute the values into the formula.

50,000 \ J=2260 \ J/g*m

We want to find the mass. We must isolate the variable, m.

m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.

\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}

\frac{50,000 \ J}{2260 \ J/g} =m

Divide. Note that the Joules (J) will cancel each other out.

\frac{50,000 \ }{2260 \ g} =m

22.1238938 \ g =m

Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.

22 \ g \approx m

The mass is about 22 grams, so choice B is correct.

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2 years ago
What does Hess's law state can be done in order to be able to react solid magnesium with oxygen gas safely (that is, without exp
aleksley [76]

Answer:

C. The reaction can be broken down and performed in steps

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To obtain MgO safely without exposing magnesium to flame, the reaction sequence shown in the image attached may be carried out. Since the enthalpy of the overall reaction is independent of the pathway between the initial and final states of the system, the sum of the enthalpy of each step yields the enthalpy of formation of MgO.

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bazaltina [42]

Answer:

24m/s

Explanation:

a=change of v/change of t

6m/s^2=v/4s

multiply both sides by 4s

v=24m/s

4 0
2 years ago
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