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3 years ago

? P+ ? Br→ ? PBT

Chemistry

1 answer:

madreJ [45]3 years ago

3 0

Answer b 2 6 2
Took test

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Nonliving things have cells.<br><br> True<br> False

Whitepunk [10]

The answer is false.

3 0

3 years ago

Read 2 more answers

What volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below

mina [271]

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces 22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

8 0

3 years ago

2CH3CHO+O2=> 2CH3COOH

Dovator [93]

Answer:

2CH3CHO + O2 → 2CH3COOH

7 0

2 years ago

Round 647.155 to nearest hundredt

lisov135 [29]

Answer:600

Explanation:

5 0

2 years ago

Match each transition metal ion with its condensed ground-state electron configuration. A [Ar]3d2 B [Ar]4s23d3 C [Kr]4d10 D [Xe]

madreJ [45]

Answer:

$Mn^{2+}$ : - F . $[Ar]3d^{5}$

$Hg^{2+}$ : - G. $[Xe]4f^{14}5d^{10}$

$La^{3+}$ : - D. $[Xe]$

$Fe^{3+}$ : - F. $[Ar]3d^{5}$

$Ag^{+}$ : - C. $[Kr]4d^{10}$

$Co^{3+}$ : - E. $[Ar]3d^{6}$

Explanation:

The electronic configuration of the element Mn is:-

$[Ar]3d^{5}4s^2$

For, $Mn^{2+}$ , 2 electrons are lost, thus the configuration is:-

$[Ar]3d^{5}$

The electronic configuration of the element Hg is:-

$[Xe]4f^{14}5d^{10}6s^2$

For, $Hg^{2+}$ , 2 electrons are lost, thus the configuration is:-

$[Xe]4f^{14}5d^{10}$

The electronic configuration of the element La is:-

$[Xe]5d^{1}6s^2$

For, $La^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Xe]$

The electronic configuration of the element Fe is:-

$[Ar]3d^{6}4s^2$

For, $Fe^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Ar]3d^{5}$

The electronic configuration of the element Ag is:-

$[Kr]4d^{10}5s^1$

For, $Ag^{+}$ , 1 electron is lost, thus the configuration is:-

$[Kr]4d^{10}$

The electronic configuration of the element Co is:-

$[Ar]3d^{7}4s^2$

For, $Co^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Ar]3d^{6}$

7 0

3 years ago