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2 years ago

If a proton is added to the nucleus of germaium what out comes would acure

1 answer:

3 0

When a proton is added to the germanium nucleus, then its atomic number becomes 33, which is the atomic number of arsenic.

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No. 65, chemistry help Thank you

dexar [7]

Remember the acronym "Oil Rig". Oxidation is loss, Reduction is gain of electrons. Calcium is losing electrons so it's an oxidation reaction.

7 0

3 years ago

Predict the geometry around the central atom in SO2- A) trigonal pyramidal B) tetrahedral C) octahedral D) trigonal planar E) tr

liubo4ka [24]

Answer: The correct answer is Option D.

Explanation:

To calculate the hybridization of $SO_2$ , we use the equation:

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom (S) = 6

N = number of monovalent atoms bonded to central atom = 0

C = charge of cation = 0

A = charge of anion = 0

Putting values in above equation, we get:

$\text{Number of electron pair}=\frac{1}{2}[6]=3$

The number of electron pair around the central metal atom are 3. This means that the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar.

Hence, the correct answer is Option D.

3 0

3 years ago

The accepted value for R is 0.08206 L·atm/K·mol or 8.314 L·kPa/K·mol, depending on the unit of pressure used. (Your answer may d

Alecsey [184]

Answer:

poop

Explanation:

8 0

3 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

2 years ago

Is h3co3 a strong base

tamaranim1 [39]

I read and said it's a weak base

3 0

3 years ago