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2 years ago

Do we need nuclear energy to stop climate change?

Chemistry

1 answer:

Rom4ik [11]2 years ago

5 0

Answer:

Hey mate......

Explanation:

This is ur answer.....

<em>To combat climate change, th</em><em>e</em><em> </em><em>world must rapidly reduce its dependency on fossil fuels to reduce greenhouse gas emissions. Nuclear energy is low-carbon and can be deployed on a large scale in the time frame required, supplying the world with clean and affordable electricity.</em>

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Seawater is around a 3% aqueous salt solution.

Eduardwww [97]

Answer:

the answer is distillation

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1 year ago

3. Hydrogen reacts with nitrogen to produce ammonia according to the equation:

lys-0071 [83]

Mass of ammonia produced : 121.38 g

<h3>Further explanation</h3>

Given

Reaction

3H₂(g) + N₂(g) ⇒ 2NH₃(g)

100g of N₂

Required

Ammonia produced

Solution

mol of N₂ :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{100}{28}\\\\mol=3.57$

From the equation, mol ratio of N₂ and NH₃ = 1 : 2, so mol NH₃ :

$\tt \dfrac{2}{1}\times 3.57=7.14~moles$

mass of NH₃(MW=17 g/mol) :

$\tt mass=mol\times MW\\\\mass=7.14\times 17\\\\mass=121.38~g$

8 0

3 years ago

What does not add co2 to the atmosphere

marissa [1.9K]

Answer:

Abiotic objects

Ex. bed, table, lamp, chair, blanket, etc.

Abiotic means Non-living.

7 0

2 years ago

2.<br> An alkane has at least on C=C<br> bond.<br> ut of<br> Select one:<br> O True<br> O False

garik1379 [7]

I think it’s false?????

8 0

3 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

3 0

3 years ago

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