All categories

2 years ago

Which of the following molecules has one lone pair of electrons?

2 answers:

7 0

You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.

jeyben [28]2 years ago

3 0

The ammonia molecule contains three single bonds and one lone pair on the central nitrogen atom . The domain geometry for a molecule with four electron pairs is tetrahedral, as was seen with CH4. In the ammonia molecule, one of the electron pairs is a lone pair rather than a bonding pair.

Hope this helps :}

-bucky

You might be interested in

Can anyone help me with 6-10 ? <br> I’ll mark as a brainliest.

Ivanshal [37]

Answer:

based on this ball and stick model how many oxygen atoms are there in a milecule of 2 propanol

Explanation:

7 0

3 years ago

Read 2 more answers

How many water molecules are need to break down an oligosaccharide that contains 10 glucose units?

mestny [16]

Answer is 10 units of water molecules.

3 0

3 years ago

How many miles of O2 can be produced by letting 12.00 miles of KC103 react?

Marat540 [252]

2KClO3 --> 2KClO2 + O2
12 6 (moles)
The ratio of KClO3 and O2 is 2:1. This means 2 moles of KClO3 can create 1 mole of O2. So 12 moles of KClO3 will create 6 moles of O2.

3 0

2 years ago

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$

Replacing every value in L/d and mg/L

$C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L$

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).

We have the total flow F3 = F1 + F2 and the final concentration $C_f$ . It is required to calculate per day, let's take a time of t = 1 day.

$F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg$

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0

3 years ago

Which term names the main class of the chemical that makes up bread.

Shkiper50 [21]

Answer:

yeast

Explanation:

4 0

3 years ago