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blagie [28]
3 years ago
5

If 1.0g of rubidium and 1.0 g of bromine are reacted what will be left in measurable amounts in the reaction

Chemistry
1 answer:
pentagon [3]3 years ago
7 0

Answer:

                     0.0639 g of Br₂ will be left

Explanation:

                    The balance chemical equation for given synthesis reaction is,

                                          2 Rb + Br₂ → 2 RbBr

Step 1: <u>Calculate Moles of each reactant for given masses as;</u>

Moles of Rb = 1 g ÷ 85.46 g/mol

Moles of Rb = 0.0117 moles

Similarly,

Moles of Br₂ = 1 g ÷ 159.80 g/mol

Moles of Br₂ = 0.00625 moles

Step 2: <u>Find reactant in Excess as:</u>

According to equation,

                    2 moles of Rb reacted with  =  1 mole of Br₂

So,

              0.0117 moles of Rb will react with  =  X moles of Br₂

Solving for X,

                      X  =  0.0117 mol × 1 mol / 2 mol

                     X =  0.00585 mol of Br₂

While, as calculated above we are provided with 0.00625 moles of Br₂. Therefore, Br₂ is in excess and the excess amount is calculated as,

= Given Moles - Consumed Moles

= 0.00625 moles - 0.00585 moles

= 0.0004 moles

Step 3: <u>Calculate mass of excess Br₂ as:</u>

Mass = Moles × M.Mass

Mass = 0.0004 mol × 159.80

Mass = 0.0639 g of Br₂

Note: Speaking broadly, there will be 1.93 g of RbBr along with 0.0639 g of Br₂ because RbBr is being produced and is available along with unreacted Br₂.

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A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the
Gnesinka [82]

Answer:

<em><u>CaAsHO₄</u></em>

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

<u>Step 1. Calculate the amount in moles of each element.</u>

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

<u>Step 3. Write the empirical formula with the rounded numbers obtained</u>

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

<em><u>CaAsHO₄</u></em>

3 0
3 years ago
9) Give the set of four quantum numbers that could represent the extra electron added (using the Aufbau principle) to the neutra
Pie

Answer:

See explanation

Explanation:

9) If an extra electron is added to the neon atom, then the electronic configuration becomes; 1s2 2s2 2p6 3s1

This last electron has quantum numbers;

n=3, l=0, m=0 and s = +1/2

This is so because the 2s level is already filled so the extra electron must go into the 3s level. The orbital quantum number and the magnetic quantum number for the s orbital is zero.

10) Electron affinity is the energy released when one mole of gaseous atoms accept one mole of gaseous electrons to form one mole of gaseous ions having a negative charge.

In the second period, fluorine has the greatest electron affinity since electron affinity increases across the period. The noble gas, neon has an electron affinity of 0KJ/mol.

11) Ionization energy decreases down the group but increases across the period due to increase in the size of the nuclear charge and decrease in the distance between the nucleus and the outermost electron. Hence, the process; Br+(g) ---->Br2+(g) + e- has the greatest ionization energy. Recall that the second ionization is always higher than the first ionization energy.

12)The order of decreasing metallic character here is K> As> P. Even though As and P belong to the same group, we must note that metallic character increases down the group hence the order written above.

3 0
2 years ago
PLZ HELP! Which of the following statements is true?
Sindrei [870]
<span>1. over thrust plate boundary
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</span>
4 0
3 years ago
When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu
olga2289 [7]

Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC  = C = 0,50 J/gºC

8 0
3 years ago
Is mass conserved when 50 g of sugar undergoes a physical change?
Lena [83]
Yes, mass never changes. No exceptions.
6 0
3 years ago
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