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3 years ago

Which atom’s ionization energy is greater than that of phosphorus (P)?

Chemistry

1 answer:

Anna71 [15]3 years ago

7 0

The first ionisation energy of silicon is greater than that of phosphorus.

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What are the total electron pairs in AsH3

tatiyna

AsH3 has five valence electrons, so the molecule will have 3 bonding domains and one non-bonding domain (lone pair) around the central atom. The geometry is trigonal pyramidal, and the angles are around 107 degrees.

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4 years ago

What is the molecular geometry of CO2?

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Answer:

Tetrahedral

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3 years ago

The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2

Kruka [31]

Hey there!

For this we can use the combined gas law:

$\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

We are only working with pressure and temperature so we can remove volume.

$\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}$

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

$\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}$

Solve for T₂.

$2atm = \frac{2.2atm}{T_{2}}*27C$

$2atm * T_{2}={2.2atm}*27C$

$T_{2}={2.2atm}\div2atm*27C$

$T_{2}=1.1*27C$

$T_{2}=29.7C$

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

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3 years ago

A student lives in a place that receives lots of snow and ice during the winter. The student has observed that the highway depar

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A. Salt lowers the freezing point of water, which makes the melted snow on the road less likely to form ice.

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4 years ago

Read 2 more answers

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$