In addition reaction ,where does the 4 electrons of double bond end up after the reaction?

In a _______ reaction, the compound breaks down; in a _______ reaction, a single compound is assembled from two or more substanc

Alexxx [7]

It's letter C. When ever you read "breaks down" always think decomposition. When you read that compound get together to form "a single" compound, think synthesis.

7 0

3 years ago

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What is the correct name for the ionic compound, MgCl2?

algol [13]

Answer:

magnesium chloride (no prefixes)

8 0

2 years ago

When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t

satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

$Q=C\times Delta T$

$Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J$

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = $\frac{15.3 g}{85 g/mol}=0.18 mol$

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

$\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ$

1 J = 0.001 kJ

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

8 0

3 years ago

This image shows a...

likoan [24]

Answer:

I think is the last one!!!! punnet square ⬜

5 0

2 years ago

Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7

11Alexandr11 [23.1K]

Explanation:

For an isothermal process equation will be as follows.

W = nRT ln $\frac{P_{1}}{P_{2}}$

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{10000 g/s}{18 g/mol}$

= 555.55 mol/s

= 556 mol/s (approx)

As T = $50^{o}C$ or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

= $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}$

= $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303$

= -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

$3440193.809 J/s \times \frac{0.001 kJ}{1 J}$

= 3440.193 kJ/s

= 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

6 0

3 years ago