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4 years ago

15

The process___helps us to calculate specific heat of unknown samples/metals OR helps us calculate changes due to energy transfer

between a system and its surrounding.
type your answer...

1 answer:

ohaa [14]4 years ago

7 0

I believe it’s calorimetry

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A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5

Marina CMI [18]

Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

1.Use Boyle's Law( $P_{1} V_{1}= P_{2} V_{2}$ ). Re-arrange to solve for $V_{2}$ <em> for the final volume.</em>

<em />

<em>2. Plug in values. </em> $V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)} = 7.14 mL$

3 0

2 years ago

Which of the following is an oxidation-reduction reaction? Fe2O3 3CO Right arrow. 2Fe 3CO2 CuSO4 2NaOH Right arrow. Cu(OH)2 Na2S

Damm [24]

The reaction, Fe2O3 + 3CO------> 2Fe + 3CO2 is an oxidation-reduction reaction.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number from left to right in the reaction. This is because, a specie is oxidized and another specie is reduced.

In the reaction; Fe2O3 + 3CO------> 2Fe + 3CO2, we can see that the oxidation number of iron decreased from +3 on the left hand side to zero on the right hand side. The oxidation number of carbon was increased from + 2 to +4.

Learn more: brainly.com/question/10079361

6 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

5 0

3 years ago

This is for my science class.

Kay [80]

Answer:

4 joules

Hope this helps if it does consider giving brainliest

Cya on the next one!

Explanation:

5 0

3 years ago

Read 2 more answers

3. A student adds 0.400g of n-propanol to 13.0 g of t-butanol. What is the molality of the solution? Show your calculations. (3

marissa [1.9K]

Answer:

$m=0.512m$

Explanation:

Hello,

In this case, we can consider the n-propanol as the solute (lower amount) and the t-butanol as the solvent (higher amount), for which, initially, we must compute the moles of n-propanol (molar mass = 60.1 g/mol) as shown below:

$n_{solute}=0.400g*\frac{1mol}{60.1g0}=6.656x10^{-3}mol$

Since the molality is computed via:

$m=\frac{n_{solute}}{m_{solvent}}$

Whereas the mass of the solvent is used in kilograms (0.0130g for the given one). Thus, we compute the resulting molality of the solution:

$m=\frac{6.656x10^{-3}mol}{0.0130kg}\\ \\m=0.512\frac{mol}{kg}$

Or just:

$m=0.512m$

Best regards.

8 0

3 years ago