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3 years ago

Does a light with low frequency and long wavelength have a lot or little energy?

1 answer:

7 0

The lower the frequncy the longer the wavelength the longer the wavelength whoch would probably give you more energy.

if im incorrect im truly sorry

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A certain metal oxide has the formula MO where M denotes the metal. A 61.16−g sample of the compound is strongly heated in an at

Reptile [31]

Answer:

the metal is Cadmium (Cd), the oxide is Cadmium oxide (CdO)

Explanation:

Step 1: The equation

mMO ⇔ mM + mO

Step 2: calculating mass

Given is the mass of the metaloxide (61.16g) and the mass of the metal, since the oxygen was removed (53.54g)

The mass of the oxide can be calculate by:

mO = mMO - mM

mO = 61.16 - 53.54 = 7.62g

Step 3: Calculating moles oxide

mole = mass/ molar mass

mole = 7.62g / 16g/mole = 0.47625 mole

Step 4: moles metal

For 1 mole O2 we will have 1 mole metal

so we have 0.47625 moles of metal

Step 5: Calculating molar mass of metal

Molar mass = mass / moles

Molar mass = 53.54g / 0.47625 mole

Molar mass = 112.41 g/mole

If we look at the periodic table ⇒ Cadmium has a molar mas of 112.41 g/mole

So the metal is Cadmium (Cd), the oxide is Cadmium oxide (CdO)

6 0

3 years ago

I don’t really have a question... and my teacher says I need to give one related to this reading above :/

irinina [24]

Answer:

Just say I wonder why teachers give homework :/

Explanation:

8 0

4 years ago

The molar concentration (M) of a solution prepared by dissolving 0.2362g of Cr(NO3)3 in a 50-mL volumetric flask is 0.01985M, wh

Vinil7 [7]

Answer:

Here's what I get

Explanation:

You want to dilute the original solution by a factor of 25 in two steps, so you could dilute it by a factor of 5 in the first step, then dilute the new solution by another factor of 5.

A. First dilution

Use a 10 mL pipet to transfer 10 mL of the original solution to a 50 mL volumetric flask. Make up to the mark with distilled water. Shake well to mix.

Use the dilution formula to calculate the new concentration.

$\begin{array}{rcl}c_{1}V_{1} & = & c_{2}V_{2}\\0.01985 \times 10.00 & = & c_{2} \times 50.00\\0.1985 & = & 50.00 c_{2}\\\\c_{2}& = & \dfrac{0.1985}{50.00}\\\\& = & \text{0.003 970 mol/L}\\\end{array}$

B. Second dilution

Repeat Step 1, using the 0.003 970 mol·L⁻¹ solution.

$\begin{array}{rcl}c_{2}V_{2} & = & c_{3}V_{3}\\0.003970 \times 10.00 & = & c_{3} \times 50.00\\0.03970 & = & 50.00 c_{3}\\\\c_{3}& = & \dfrac{0.03970}{50.00}\\\\& = & \textbf{0.000 7940 mol/L}\\\end{array}\\\text{The concentration of the final solution is $\boxed{\textbf{0.000 7940 mol/L}}$}$

3. Check:

Compare the final concentration with the original

$\begin{array}{rcl}\dfrac{ c_{3}}{ c_{1}} & = & \dfrac{0.0007940}{0.01985}\\& = & \mathbf{\dfrac{1}{25.00}}\\\end{array}\\\text{The concentration of the final solution is } \boxed{\mathbf{\dfrac{1}{25}}} \text{ that of the original solution}$

7 0

3 years ago

A 2.600×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w

sesenic [268]

Answer:

A. 0.2395 w/w %

B. 2394ppm

Explanation:

A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:

Mass glycerol / Total mass * 100

Mass glycerol:

The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):

2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol

Mass of water:

998.9mL and density = 0.9982g/mL:

998.9mL * (0.9982g/mL) = 997.1g of water.

That means percent by mass is:

% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %

B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:

2.394g * (1000mg / 1g) = 2394mg:

Parts per million: 2394mg / L = 2394ppm

3 0

3 years ago

Consider the balanced equation.

viktelen [127]

28.4% is the answer and that is ur answer

8 0

2 years ago