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4 years ago

Both active solar heating systems and passive solar heating systems

Chemistry

2 answers:

s2008m [1.1K]4 years ago

5 0

A. reduce the need to use fossil fuels

Arada [10]4 years ago

4 0

Answer:

The answer is A) REDUCE THE NEED TO USE FOSSILS FUELS GRADPOINT

GOOD LUCK!

Explanation:

You might be interested in

Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 9

Natali5045456 [20]

Answer:

– 844 kJ/mol.

Explanation:

The following data were obtained from the question:

N2(g) + 3 F2(g) –––> 2 NF3(g)

Enthalpy of N≡N (N2) = 945 kJ/mol

Enthalpy of F–F (F2) = 155 kJ/mol

Enthalpy of N–F3 (NF3) = 283 kJ/mol

Enthalpy change (∆H) =?

Next, we shall determine the enthalpy of reactant.

This is illustrated below:

Enthalpy of reactant (Hr) = 945 + 3(155)

Enthalpy of reactant (Hr) = 945 + 465

Enthalpy of reactant (Hr) = 1410 kJ/mol

Next, we shall determine the enthalpy of the product.

This is illustrated below:

Enthalpy of product (Hp) = 2 x 283

Enthalpy of product (Hp) = 566 kJ/mol

Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:

Enthalpy of reactant (Hr) = 1410 kJ/mol

Enthalpy of product (Hp) = 566 kJ/mol

Enthalpy change (∆H) =?

Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)

Enthalpy change (∆H) = 566 – 1410

Enthalpy change (∆H) = – 844 kJ/mol

7 0

3 years ago

1. An electric power plant uses energy from burning coal to generate steam at 450 °C. The plant is cooled by 20 °C water

kap26 [50]

Answer:

Explanation:

Efficiency of the electric power plant is $e=1-\frac{T_{2}}{T_{1}}$

Here Temperature of hot source $T_{1} = 450^{o}C=450+273=723 K$

and Temperature of sink $T_{1} = 20^{o}C=20+273=293 K$

Hence the efficiency is $e=1-\frac{293}{723}=0.5947=59.47%$

Now another formula for thermal efficiency Is

$e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}$

Here QI is the of heat taken from source 100 MJ ; Q2 of heat transferred to the sink (river) to be found

W is the of work done and W = QI -Q2

Hence From $e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}$

$W=e(Q_{1})=(0.5947)(100)=59.47MJ$

Hence the of heat transferred to the river Is $Q_{2} -W = (100 -59.47=40.53$

6 0

4 years ago

Please begging you guys someone help me

BaLLatris [955]

Answer:

See Explanation

Explanation:

10) From the options provided for this question, gamma particle is the most energetic. Recall that gamma rays are high energy electromagnetic radiation which are capable of causing a high degree of ionization in matter.

11) The bombardment of U-235 with neutrons leads to the reaction;

$U\frac{235}{92} + n\frac{1}{0}---> I\frac{138}{53} + Y\frac{95}{39} +3n \frac{1}{0}$

Hence

a = 92, b= 95, c= 53

12) In positron emission, a proton is transformed into a neutron. The mass number of the daughter nucleus is the same as its parent but the atomic number decreases by 1.

Hence;

$Th\frac{231}{90} -----> e\frac{1}{0} +Ac \frac{231}{89}$