Helppp its qboutt cellsss

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0

3 years ago

How many atoms of carbon are in a container that has 0.450 mol of CO2 and 2.55 mol of CaC2?

Fantom [35]

Answer:

The number of carbon atoms in the container is 1.806 × 10²⁴ or the container contains 1.806 × 10²⁴ atoms of carbon

Explanation:

By Avogadro's number, 1 mole of a substance contains 6.02 × 10²³ particles of the substance

Here we have 0.45 mole of CO₂ contains

0.45 × 6.02 × 10²³ particles of CO₂ that is 2.709 × 10²³ particles of CO₂ or equivalent to 2.709 × 10²³ atoms of Carbon

Similarly, 2.55 moles of CaC₂ contains 2.55 × 6.02 × 10²³ particles of CaC₂ or 1.5351 × 10²⁴ atoms of Carbon

The total number of carbon atoms is therefore;

2.709 × 10²³ + 1.5351 × 10²⁴ = 1.806 × 10²⁴ atoms of carbon.

5 0

3 years ago

An example of a disaccharide is saccharin-aspartame molecule True False

Flura [38]

I think is False! Because it is not example of a disaccharide it is not saccharin-aspartame molecule.

6 0

2 years ago

Calculate the ph of a 0.17 m solution of c6h5nh3no3 (kb for c6h5nh2 = 3.8 x 10-10). record your ph value to 2 decimal places.

MatroZZZ [7]

<span>Mass of the solution = 0.17m
Kb for C6H5NH2 = 3.8 x 10^-10
We know Ka for C6H5NH2 = 1.78x10^-11
We have Kw = Ka x Kb => Ka = Kw / Kb
=> (C2H5NH2)(H3O^+)/(C2H5NH3^+) => 1.78x10^-11 = K^2 / 0.17
K^2 = 3 x 10^-12 => K = 1.73 x 10^-6.
pH = -log(Kw(H3O^+)) = -log(1.73 x 10^-6) = 5.76</span>

6 0

2 years ago

What are the four rules for significant figures

kkurt [141]

Answer:

Significant Figures

Explanation:

1) Annotation category: ...

2)RULES FOR SIGNIFICANT FIGURES.

3) All non-zero numbers ARE significant. ...

4)Zeros between two non-zero digits ARE significant. ...

5)Leading zeros are NOT significant. ...

6)Trailing zeros to the right of the decimal ARE significant. ...

7)Trailing zeros in a whole number with the decimal shown ARE significant. 8) to make you happy :) hope this helps you :)

5 0

2 years ago