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3 years ago

This is due at 2:30 i need helppppp

Chemistry

2 answers:

Liono4ka [1.6K]3 years ago

7 0

Carbon dioxide and water based on the pic:)

Mrrafil [7]3 years ago

3 0

Answer:

Oxygen and Glucose

Explanation:

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A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

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

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

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

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0

4 years ago

The ancient false idea that life appeared from nowhere

Korolek [52]

The answer is God because if you think on it hard enough, you realize that we couldnt of come from nowhere. Then, you get a headache and just stop thinking about.

Hope this helps :D

8 0

3 years ago

Read 2 more answers

What is correct about solubility of<br> substances?

Svet_ta [14]

I think it should be A

7 0

3 years ago

The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. T

olga55 [171]

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

$ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})$

And as Half-life in a first order reaction is:

$t_{1/2}=\frac{ln2}{K}$

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

$58.0min=\frac{ln2}{K}$

K = 0.01195min⁻¹ = K₁

$ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})$

$ln\frac{0.01195min^{-1}}{K_2} =1.47$

$\frac{0.01195min^{-1}}{K_2} =4.35$

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

$t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}$

5 0

3 years ago

Mwehehe poide bang pakiss?

Contact [7]

Answer:

what confused

Explanation:

6 0

3 years ago

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