If you find that a substance turns litmus paper to blue and contain hydroxide ions,it is most likely

What is the mass number of a carbon atom which has 7 neutrons?

Strike441 [17]

Answer:

Carbon-13

Explanation:

Carbon have three isotopes. Isotopes are the atoms of the same element which has a different number of neutrons. Carbon has 3 isotopes.

Carbon-12 : 6 electrons I 6 protons I 6 neutrons

Carbon-13 : 6 electrons I 6 protons I 7 neutrons

Carbon-14 : 6 electrons I 6 protons I 8 neutrons

5 0

3 years ago

What type of change occurs when rust on a nail a. chemical change b. physical change

Korvikt [17]

Answer:

Chemical change

Explanation:

The rusting of iron/a nail is a chemical change

Iron (Fe) and Oxygen (O) combine to create the compound Iron Oxide (Fe2O3), which is rust.

5 0

3 years ago

A certain photon has a frequency of 634x1012 Hz. What is the photon's wavelength in a vacuum? Enter your answer in nm (x10-9 m).

Shtirlitz [24]

<h3>Answer:</h3>

4.73 × 10^4 m

<h3>Explanation:</h3>

From the question;

Frequency of the photon = 634 × 10^12 Hz

We are required to calculate the wavelength of the photon.

We need to know the relationship between wavelength and frequency of a wave.

The relationship between f and λ is given by;

c = fλ

Where c, is the speed of light, 2.998 × 10^8 m/s

Therefore, to get the wavelength we rearrange the formula such that;

λ = c ÷ f

= 2.998 × 10^8 m/s ÷ 634 × 10^12 Hz

= 4.73 × 10^-5 m

But we require wavelength in nm

1 M = 10^9 nm

Therefore;

Wavelength = 4.73 × 10^-5 m × 10^9 nm/m

= 4.73 × 10^4 m

Hence, the photon's wavelength is 4.73 × 10^4 m

4 0

3 years ago

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA requ

egoroff_w [7]

Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

Step 1: Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

Step 2: Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+ in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed in the back titration with Ca2+:

Step 3: Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

Step 4: Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

Step 5: Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

Step 6: Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

Step 7: Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

4 0

3 years ago

You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00

Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ $\frac{[Ac^-]}{[Acac]}$

Where Ac⁻ is the salt of acetic acid (Acac).

Solving:

1,738 = $\frac{[Ac^-]}{[Acac]}$ (1)

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] (2)

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = 0,0365 mol

Ac⁻ = 0,127 M × 0,500 L = 0,0635 mol

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × $\frac{60,05 g}{1 mol}$ = 6,005 g

You need to weight 6,005 g of acetic acid

I hope it helps!

7 0

4 years ago