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Sever21 [200]
3 years ago
5

g How many grams of beryllium are needed to produce "11.5" g of hydrogen gas?Be(s) + 2H 2 O(l) → Be(OH) 2 (aq) + H 2 (g)

Chemistry
1 answer:
slega [8]3 years ago
6 0

Answer:

m_{Be}=51.3gBe

Explanation:

Hello,

In this case, with the given chemical reaction:

Be(s) + 2H_2 O(l) \rightarrow Be(OH)_ 2 (aq) + H _2 (g)

We can notice a 1:1 mole ratio between hydrogen gas and beryllium metal, therefore, beryllium requirement is computed as shown below:

m_{Be}=11.5gH_2*\frac{1molH_2}{2.02gH_2} *\frac{1molBe}{1molH_2} *\frac{9.01gBe}{1molBe}\\ \\m_{Be}=51.3gBe

Regards.

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Weight of one mole of carbon = 12.01 g Weight of one mole of oxygen = 16.00 g The molecular weight (gram formula weight) for CO
murzikaleks [220]

Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .

The molecular weight of the compound CO is therefore

equal to the sum of the weight of both elements.

That’s = 12.01g + 16.00g

= 28.01g

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4 0
3 years ago
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is
Airida [17]

Answer:

the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C  

Explanation:

Given the data in the question;

First we calculate the Volume of the steel cylinder;

V = πr²h

radius r = Diameter / 2 = 27 cm / 2 = 13.5 cm

height h = 32.4 cm

so we substitute

V = π × ( 13.5 cm )² × 32.4 cm

V  = π × 182.25 cm × 32.4 cm

V = 18550.79 cm³  

V = 18.551 L

given that; maximum safe pressure P = 3.10 MPa = 30.5946 atm

vessel contains 0.218kg or 218 gram of carbon monoxide gas

molar mass of carbon monoxide gas is 28.010 g/mol

so

moles of carbon monoxide gas n = 218 gram /  28.010 g/mol = 7.7829 mol

we know that;

PV = nRT

solve for T

T = PV / nR

we know that gas constant R = 0.0820574 L•atm•mol⁻¹ K⁻¹

so we substitute

T = ( 30.5946 × 18.551 ) / ( 7.7829 × 0.082 )

T = 567.5604 / 0.6381978

T = 889.317387 K

T = ( 889.317387 - 273.15 ) °C

T = 616.167 ≈ 616 °C  { 3 significant digits }

Therefore, the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C  

6 0
2 years ago
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