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Ilia_Sergeevich [38]
3 years ago
8

Carbon containing compounds are referred to as:

Chemistry
1 answer:
Flura [38]3 years ago
3 0
Organic compounds
If you need an explanation I can give one!
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What is the combination of atoms Formed by a covalent bond
zysi [14]

Answer:

Explanation:

In a polar covalent bond, the distribution of common electrons are not shared evenly due to a greater positive charge from one atom's nucleus.Oct 30, 2016

3 0
3 years ago
In a recent Grand Prix, the winner completed the race with an average speed of 299.8 km/h. What was his speed in miles per hour,
scoundrel [369]

Answer :

(a) The speed is, 186.287 mi/hr

(b) The speed is, 83.277 m/s

(c) The speed is, 273.22 feet/s

Explanation :

(a) We are given the speed is, 299.8km/hr

As we know that,

1km=\frac{1}{1.60934}mi

By using both the conversion, we get:

299.8km/hr

\Rightarrow \frac{299.8km}{1hr}\times \frac{1mi}{1.60934km}

\Rightarrow 186.287mi/hr

The speed is, 186.287 mi/hr

(b) We are given the speed is, 299.8km/hr

As we know that,

1km=1000m

and,

1hr=3600s

By using both the conversion, we get:

299.8km/hr

\Rightarrow \frac{299.8km}{1hr}\times \frac{1000m}{1km}\times \frac{1hr}{3600s}

\Rightarrow 83.277m/s

The speed is, 83.277 m/s

(c) We are given the speed is, 299.8km/hr

As we know that,

1km=3280.84feet

and,

1hr=3600s

By using both the conversion, we get:

299.8km/hr

\Rightarrow \frac{299.8km}{1hr}\times \frac{3280.84feet}{1km}\times \frac{1hr}{3600s}

\Rightarrow 273.22feet/s

The speed is, 273.22 feet/s

3 0
4 years ago
Which term describes a lens with a surface that curves outward like the exterior of a sphere
ss7ja [257]
The answer would be D.) convex
8 0
3 years ago
4.14 Oxygen requirement for growth on glycerol Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as
Iteru [2.4K]

Answer:

0.37 g

Explanation:

The molecular weight for Glycerol = 92

Number of Carbon atoms in glycerol (x)  C_3H_8O_3 = 3

Molecular weight of  the Biomass  ( Klebsiella aerogenes )

= CH_{1.73}O_{0.43}N_{0.24}

= \frac{23.97}{0.92}

= 26.1

From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as \delta _g) as follows:

= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)

= 4.15

Given that the yield of the Biomass = 0.40 g

However;

C = Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}

C = 0.40*\frac{92}{26.1}

C = 1.41 g

Now , the oxygen requirement can be calculated as:

= \frac{1}{4}*(n*S -  C * \delta _{g})

= \frac{1}{4}(3*4.7-1.41*4.15)

= 2.1 g/mol

Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.

Now converting it to mass terms; we have:

= 2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}

= 2.1 * \frac{16}{92}

= 0.3652 g

≅ 0.37 g

∴ The oxygen requirement for this culture in mass terms = 0.37 g

3 0
3 years ago
Please help with the last 2.
ASHA 777 [7]

Answer: Yttruim: 39 protons, 50 neutrons, 39 electrons

Ruthenium: 44 protons, 57 neutrons, 44 electrons

3 0
3 years ago
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