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erastovalidia [21]

4 years ago

15

1. How does adding energy to a solid affect the motion of the particles? 2. How can a gas at room temperature (like oxygen) beco

me solid?

1 answer:

Oksanka [162]4 years ago

6 0

I am not sure sorry

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Why does Ca have a lower ionization energy than He <br> ill give u brainliest :)

sleet_krkn [62]

Answer: because he has more protons than ca

Explanation:

7 0

3 years ago

A wave has a frequency of 3,5MHz and a wavelength of 85,5 m. What is the wave velocity through

Yuki888 [10]

Answer:

velocity = 29925×10⁶ m/s

Explanation:

Given data:

Frequency = 35 MHz

Wavelength = 855 m

Velocity = ?

Solution:

MHz to Hz

35×10⁶ Hz

Formula:

<em>velocity = Wavelength × Frequency</em>

velocity = 855 m × 35×10⁶ Hz

velocity = 29925×10⁶ m/s

Hz = s⁻¹

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3 years ago

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Elanso [62]

Answer:

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4) The tetanus vaccine is made by taking the tetanus toxin and inactivating it with a chemical. The inactivated toxin is called a "toxoid." Once injected, the toxoid elicits an immune response against the toxin

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8)help your immune system protect against these infections

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7 0

3 years ago

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What happens when resource depletion occurs

sladkih [1.3K]

A large fraction of a specific source is used up.

5 0

3 years ago

Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:

Sophie [7]

Answer : The enthalpy change for the reaction is, 97.7 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main chemical reaction is,

$2C(s)+3H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=131.3kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=41.2kJ$

(3) $CH_4(g)+H_2O(g)\rightarrow 2H_2(g)+CO(g)$ $\Delta H_3=206.1kJ$

Now we are multiplying reaction 1 by 2 and reversing reaction 3 and then adding all the equations, we get :

(1) $2C(s)+2H_2O(g)\rightarrow 2CO(g)+2H_2(g)$ $\Delta H_1=2\times 131.3kJ=262.6kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=41.2kJ$

(3) $2H_2(g)+CO(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206.1kJ$

The expression for enthalpy of main reaction will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(262.6)+(41.2)+(-206.1)$

$\Delta H=97.7kJ$

Therefore, the enthalpy change for the reaction is, 97.7 kJ

3 0

4 years ago

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