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4 years ago

In which liquid does solid brass float?

Chemistry

2 answers:

zimovet [89]4 years ago

7 0

It’s A , Liquid Mercury

ch4aika [34]4 years ago

6 0

Answer:

Explanation:

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How many moles are in 54.73 L of Helium?

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The answer is 13.67 moles

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3 years ago

Choose all the answers that apply.

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Answer:

All of the above. The CV system transports blood and plasma the do all 4.

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3 years ago

Which gas occupies the highest volume at STP?

Vedmedyk [2.9K]

The gas, 2 mol of H2, occupies the highest volume at STP since at STP the volume of this gas is approximately 44.8 mol as compared to other options this has the greatest amount.

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4 years ago

Read 2 more answers

What types of R groups can exist in amino acids? (Choose all that apply)

PilotLPTM [1.2K]

Answer:

acidic,non polar,basic

Explanation:

that’s the ones I put and I got a 100% on my test just now

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3 years ago

A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ris

Blababa [14]

<u>Answer:</u> The heat capacity of calorimeter is $15.66J/^oC$

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

$\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released = ?

n = number of moles = 0.473 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -1235 kJ/mol = $-1235\times 10^3J/mol$ (Conversion factor: 1 kJ = 1000 J)

Putting values in above equation, we get:

$-1235\times 10^3J/mol=\frac{q}{0.473mol}\\\\q=(-1235\times 10^3J/mol\times 0.473mol)=-584.16\times 10^3J$

To calculate the heat capacity of calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed by the calorimeter = $584.16\times 10^3J$

c = heat capacity = ?

$\Delta T$ = change in temperature = $T_2-T_1=62.3^oC-25^oC=37.3^oC$

Putting values in above equation, we get:

$584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC$

Hence, the heat capacity of calorimeter is $15.66J/^oC$

4 0

4 years ago