Which elements are metalloids ? Check all that apply

How do SMRs provide more flexibility than ordinary nuclear reactors?

Shkiper50 [21]

Answer:

The SMRs can adjust their energy output using coolants and control rods.

Power plants can be built to fit locations with less demand or less space.

Explanation:

I think that it is but I am not sure.

7 0

2 years ago

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g) In a titration experiment, H2O2(aq) reacts with aqueous Mn

Tomtit [17]

Answer:

Oxygen in hydrogen peroxide oxidizes from -1 to 0.

Explanation:

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

$5 H_2O_2_{(aq)} + 2 MnO_4^-_{(aq)} + 6 H^+_{(aq)}\rightarrow 2 Mn^{2+}_{(aq)} + 8 H_2O_{(l)} + 5 O_2_{(g)}$

Manganese in $MnO_4^-$ has oxidation state of +7

Manganese in $Mn^{2+}$ has an oxidation state of +2

It reduces from +7 to +2

Oxygen in hydrogen peroxide has an oxidation state of -1.

Oxygen in molecular oxygen has an oxidation of 0.

Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.

8 0

2 years ago

50. The Sun's energy and composition is provided by which of the following?

PIT_PIT [208]

Answer:

A. the burning of fossil fuels within the Sun

Explanation:

3 0

2 years ago

Who wants a metal?

tiny-mole [99]

Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?

A. Both have the same molar mass.

B. Both have the same number of ions.

C. Both are made up of 6.02 mc014-1 1023 molecules.

D. Both are made up of 6.02 mc014-2 1023 formula units.

The correct answer on E.D.G is ---Both are made up of 6.02 mc014-2 1023 formula units. D

8 0

3 years ago

Read 2 more answers

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

3 years ago