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3 years ago

How much Sr(OH)2 • 8 H2O (M = 265.76) is needed to

1 answer:

3 0

First, you need to find the number of moles of OH⁻ in a 250mL solution of 0.100M OH⁻. To do this, multiply 0.250L by 0.100M to get 0.025mol OH⁻. Then you use the fact that 1 mole of Sr(OH₂)·8H₂O contains 2 moles of OH⁻ which means that 0.0125mol of Sr(OH)₂·8H₂O contains 0.025mol OH⁻ (0.025/2=0.0125). Then to find the amount of Sr(OH)₂·8H₂O is needed you multiply its molar mass (265.76g/mol) by 0.0125mol to get 3.322g.
Therefore you need 3.322g of Sr(OH)₂·8H₂O.

I hope that helps. Let me know if anything is unclear.

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The frequency of an x-ray wave is 3.0 x 1012MHz. Its wave speed is 3.0x 108m/s. Calculate the wavelength of the x-ray wave below

jasenka [17]

Answer: The wavelength of the x-ray wave is $10^{-10}m$

Explanation:

To calculate the wavelength of light, we use the equation:

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

where,

$\lambda$ = wavelength of the light = ?

c = speed of x-ray= $3.0\times 10^8m/s$

$\nu$ = frequency of x-ray = $3.0\times 10^{12}MHz=3.0\times 10^{18}Hz= 3.0\times 10^{18}s^{-1}$ $(1Hz=1s^{-1})$

Putting in the values:

$\lambda=\frac{3.0\times 10^8m/s}{3.0\times 10^{18}s^{-1}}=10^{-10}m$

Thus the wavelength of the x-ray wave is $10^{-10}m$

5 0

3 years ago

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago

Draw the aromatic compound toluene (methylbenzene). show all hydrogen atoms, including those on the ring.

Otrada [13]

Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…

Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore

6= ( 4n+2)π

4 = 4n

n =1

Hence it is an aromatic compound