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3 years ago

Is anyone having a meeting only for studies?

Chemistry

2 answers:

Iteru [2.4K]3 years ago

6 0

Answer:

no sorry tho but i think i can make 1 i dont know

Explanation:

koban [17]3 years ago

4 0

Answer:yes

Explanation:

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If the neutralized mixture was boiled dry, what would be left?

alisha [4.7K]

A salt would be left over. For example, if you reacted NaOH with HCl, your products would be NaCl, a salt, and water.

7 0

3 years ago

In an experiment to determine the effect of different brands of fertilizer on the rate of plant growth, what are 4 variables tha

kherson [118]

The variables to control in an experiment to determine the effect of different fertilizers on the rate of plant growth include soil composition, temperature, water, and light.

<h3>What are controlled variables?</h3>

The expression controlled variables makes reference to experimental conditions that must be equal or constant between experimental groups in order to obtain better comparisons when collecting results.

In conclusion, The variables to control in an experiment to determine the effect of different fertilizers on the rate of plant growth include soil composition, temperature, water, and light.

Learn more about controlled variables here:

brainly.com/question/17328868

#SPJ1

5 0

2 years ago

Use the reaction: 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g) + 5720 kJ. In the reaction, 200.0g of hydrocarbon react with

Sophie [7]

Answer:

9842kJ of energy are released

Explanation:

Based on the reaction:

2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂(g) + 10 H₂O (g) + 5720 kJ

When 2 moles of C₄H₁₀ react with 13 moles of O₂ there are released 5720 kJ of energy

As molar mass of C₄H₁₀ is 58.12g/mol, moles in 200.0g of the hydrocarbon are:

200.0g C₄H₁₀ ₓ (1mol / 58.12g) = 3.441 moles of C₄H₁₀

As 2 moles of C₄H₁₀ release 5720kJ of energy, 3.441 moles of C₄H₁₀ release:

3.441 moles C₄H₁₀ ₓ (5720kJ / 2 moles C₄H₁₀) = 9842kJ of energy are released

8 0

3 years ago

Scientific Method Vocabulary To notice similarities or differences

algol [13]

Answer:

The first step would be searching for your scientific vocabulary the second step is to write it down and then that would be your answer.

Explanation:

Hope this helps!

8 0

2 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

The formula used for isothermally irreversible expansion is :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

The expression used for work done in reversible isothermal expansion will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago