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3 years ago

What are negative charged particles orbiting the nucleus called?

Chemistry

1 answer:

Helga [31]3 years ago

4 0

Answer:

Neutrons!

Explanation:

Protons are the positively charged particles while neutrons are negatively charged.

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______ is the use of enzymes is necessary to increase the activation energy requirements of a chemical reaction.

Masja [62]

Answer:

Catalysts

Explanation:

Catalysts lower the aviation energy.

6 0

3 years ago

When atoms gain electrons, they can become larger, because the addition of an electron increases electrostatic repulsion.

astra-53 [7]

Answer:True

Explanation: An anion has a larger radius than a neutral atom because it gains valence electrons. There are added electron/electron repulsions in the valence shell that expand the size of the electron cloud, which results in a larger radius for the anion.

hit the crown for me pls :)

Have a great day

6 0

2 years ago

How many molecules are in 2.00 moles of H20????????

Elis [28]

2x 6.022x10^23= 1.204x10^24

7 0

3 years ago

1.8 mol HCl and 3.3 mol NaOH react accord- ing to the equation HCl + NaOH −→ NaCl + H2O . Calculate the amount in moles of NaCl

ollegr [7]

Answer:

1.8 moles of NaCl must be produced.

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H2O

1 mol of HCl reacts with 1 mol of NaOH to produce 1mol of NaCl

To solve this question we must find, as first, the limiting reactant:

1.8 moles of HCl will need 1.8 moles of NaOH for a complete reaction (Ratio of reaction 1:1). As there are 3.3 moles of NaOH,

HCl is limiting reactant

When the 1.8 moles of HCl react completely,

1.8 moles of NaCl must be produced because 1 mole of HCl produce 1 mole of NaCl

5 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$