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3 years ago

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What is the balanced chemical equation for the reaction used to calculate ΔH∘f of SrCO3(s)? If fractional coefficients are requi

red, enter them as a fraction (i.e. 1/3). Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively without indicating allotropes. Use (aq) for aqueous solution. Express your answer as a chemical equation. View Available Hint(s)

1 answer:

Pani-rosa [81]3 years ago

6 0

Answer:

Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)

Explanation:

The standard enthalpy of formation (ΔH°f) is the energy involved in the formation of 1 mole of a substance from its elements in their most stable states. The chemical equation for the formation of SrCO₃(s) is the following.

Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)

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Suppose a thermometer has marks at every one degree increment and the mercury level on the thermometer is exactly between the 25

alekssr [168]

Answer: 25.5°C

Explanation: take the average of the reading i.e (25 + 26)/2= 25.5

5 0

3 years ago

Need help !!!!! ASAP

Julli [10]

<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

$V_{2}=1L$

<h2>Why?</h2>

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

$P_{1}V_{1}=P_{2}V_{2}$

So, we are given the information:

$V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa$

Then, isolating the new volume and substituting into the equation, we have:

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\frac{P_{1}V_{1}}{P_{2}}$

$V_{2}=\frac{50kPa*2L}{100kPa}=1L$

Hence, the new volume will be 1 L.

$V_{2}=1L$

Have a nice day!

4 0

3 years ago

Use the drop-down menus to identify the experimental group with additional nutrients and the control group without additional nu

Paladinen [302]

Answer:

A. 1 to 16

B. 17 to 32

Explanation:

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2 years ago

Read 2 more answers

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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3 years ago

1) How many Joules of energy are released when 75g of water is heated

Pie

Answer: 23,199 J

Explanation:

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2 years ago