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2 years ago

1. Copper (__) is an element on the periodic table.

Chemistry

1 answer:

user100 [1]2 years ago

3 0

Answer: Cu

Explanation: It is Cu because the origin of the word Copper comes from the latin word "Cuprum".

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defon

1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!

2) Mass fraction of this is excessive data.

3) The solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g

4 0

2 years ago

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A roll of tape measures 45.5 inches. What is the length of the tape in meters?

FinnZ [79.3K]

1.5 metres is the length of the tape. Hope this helps :)

8 0

2 years ago

How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is $PbCrO_{4}$ and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the values,

$n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol$

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0

2 years ago

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

Select all of the correct statements pertaining to the first and second ionization energies of different elements. check all tha

sertanlavr [38]

The First Ionization energy of Nitrogen is greater (Not smaller)than that of Phosphorous. This is because going down the group (N and P are in same group) the number of shells increases, the distance of valence electrons from Nucleus increases and hence due to less interaction between nucleus and valence electrons it becomes easy to knock out the electron.

<span>The second ionization energy of Na is larger than that of Mg because after first loss of electron Na has gained Noble Gas Configuration (Stable Configuration) and now requires greater energy to loose both second electron and Noble Gas Configuration. While Mg after second ionization attains Noble Gas Configuration hence it prices less energy.</span>

8 0

2 years ago

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