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GrogVix [38]
3 years ago
5

Enter an equation for the formation of NO2(g) from its elements in their standard states. Express your answer as a chemical equa

tion. Identify all of the phases in your answer.

Chemistry
1 answer:
kkurt [141]3 years ago
7 0

The standard state of the elements Nitrogen and Oxygen are N2 and O2, knowing that they are diatomic elements. With that piece of information, the unbalanced equation for the formulation of NO2(g) should be as follows -

N2 + O2 ---> NO2

And if you include their states -

N2 ( g ) + O2 ( g ) ---> NO2 ( g )

To balance this chemical equation consider the number of reactants and products on other side of the equation. If you were to include a coefficient of one - half with respect to N2 on the reactant side, it would balance the reactants and products -

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How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
3 years ago
What is the linkage that binds one monosaccharide to another?
ra1l [238]

Answer:

hope this helps

Explanation:

glycosidic bond

A covalent bond formed between a carbohydrate molecule and another molecule (in this case, between two monosaccharides) is known as a glycosidic bond (Figure 4). Glycosidic bonds (also called glycosidic linkages) can be of the alpha or the beta type.

4 0
3 years ago
Read 2 more answers
Please Help me on this question
kirza4 [7]
Acids: taste sour, has a pH less than 7
bases: tastes bitter, has a pH greater than 7
6 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
HELP . which kind of tides causes the highest and lowest tides ?
katen-ka-za [31]
Answer: spring tides
When earth, moon, and sun line up cause high and low tides :)
6 0
2 years ago
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