Ask question

Ask question

All categories

geniusboy [140]

2 years ago

10

Which of the following is NOT an example of covalent bonding? * 2 points H20 NaF NH3 O2

1 answer:

Pachacha [2.7K]2 years ago

4 0

Answer:

I THINK NH3

Explanation:

You might be interested in

How do the circulatory system and respiratory system interact and work together?

g100num [7]

I would say because the respiratory system brings in the oxygen for the body and the circulatory system moves it throughout the body with the blood

8 0

2 years ago

Read 2 more answers

I need help solving a kinetics equation.

Lady bird [3.3K]

Answer:

what would the question be?

7 0

3 years ago

Write word and balanced chemical equations to show the

Alex777 [14]

Answer:

chemical laboratory Pune Maharashtra India

5 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

2 years ago

1(a). What is the solubility of calcium chloride (CaCl2) at 5°C?

In-s [12.5K]

Answer:

60 g/100 g water

Explanation:

Find 5 °C on the horizontal axis.

Draw a line vertically from that point until you reach the solubility curve for CaCl₂.

Then draw a horizontal line from there to the vertical axis.

The solubility of CaCl₂ is 60 g/100 g water.

7 0

3 years ago