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3 years ago

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Slaked lime (Ca(OH)2) is formed from quicklime (CaO) by the addition of water. What mass of slaked lime can be produced from a m

ixture of 22.4g of CAO and 10.9g of H2O CaO(s)+H2O(l)=Ca(OH)2(s)

1 answer:

vodomira [7]3 years ago

8 0

I hope this can help you

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Determine the volume of fluid in the graduated cylinder 12.8ml 12.7ml 12.78ml 12.88

liubo4ka [24]

Answer:

the volume is 12.79ml

Explanation:

12.8+12.7+12.78+12.88=51.16

51.16/4

=12.79

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3 years ago

Question 4 (1 point)<br> If you have exactly one million atoms how many moles do you have?

rjkz [21]

Answer:

1.66 × 10⁻¹⁸ Moles

Explanation:

As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Atoms and Avogadro's Number is given as,

Number of Moles = Number of Atoms ÷ 6.022 × 10²³ Atoms/mol

Putting values,

Number of Moles = 1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

Number of Moles = 1.66 × 10⁻¹⁸ Moles

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4 years ago

An astronomer studying planets outside our solar system has analyzed the atmospheres of four planets. Which of these planets’ at

Nata [24]

Answer: Planet A: 76% Nitrogen, 23% Oxygen, 1% Other

Explanation: Hope this helps!

3 0

3 years ago

Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con

arsen [322]

Answer:

a) $T_b=590.775k$

b) $W_t=1.08*10^4J$

d) $Q=3.778*10^4J$

d) $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

$t = 293k$

Initial heat $Q=1.36 * 10^4 J$

a)

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

c)

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

d)

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

3 0

3 years ago

Số oxi hoá của nitơ trong NH3, H2SO4, NO3 lần lượt là:

NikAS [45]

Answer:

3,6,6...................................

6 0

3 years ago