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3 years ago

The elements of a group have the same number of

Chemistry

1 answer:

kogti [31]3 years ago

8 0

Answer:

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Explanation:

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A buffer solution contains 0.20 mol of propionic acid (CH3CH2COOH) and 0.25 mol of sodium propionate (CH3CH2COONa) in 1.50 dm3.

const2013 [10]

Answer:

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3 years ago

Which SI unit would be most appropriate for expressing the mass of this animal

Ede4ka [16]

Answer:

Kilograms, I think.

Explanation:

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2 years ago

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

Why is combustion of fuel necessary for heat engines to do work?

azamat

Because during combustion reaction, heat energy is released and it's this energy that is converted to work

6 0

3 years ago

Read 2 more answers

Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m

faust18 [17]

Answer:

56°

Explanation:

First calculate $a:$

$a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}$

The interplanar spacing can be calculated from:

$d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}$

The diffraction angle is determined from:

$\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476$

Solve for $\theta$

$\theta=\sin ^{-1}(0.476)=28^{\circ}$

The diffraction angle is:

$2 \theta=2\left(28^{\circ}\right)=56^{\circ}$

4 0

3 years ago