Ask question

Ask question

All categories

gizmo_the_mogwai [7]

4 years ago

12

True or false: reactants and products can be elements or compounds

1 answer:

Montano1993 [528]4 years ago

4 0

The answer is true

both reactants and products can be elementsvor compounds

You might be interested in

The isotope carbon-14 has 6 protons and an atomic mass of 14. Carbon-14 has:

Aleks [24]

Carbon has 7 neutrons

Hope this helps

5 0

3 years ago

Read 2 more answers

Pls answer this I will mark u as the brainliest...what is the formula of sodiam carbonate

kodGreya [7K]

Explanation:

Sodium Carbonate = Na2CO3

Hope it helps ya

8 0

3 years ago

Read 2 more answers

Can you help me plz I think it’s b or c

Anettt [7]

i think its C pls dont get mad if it is wrong

4 0

3 years ago

Read 2 more answers

A 1.00 L volume of HCl reacted completely with 2.00 L of 1.50 M Ca(OH)2 according to the balanced chemical equation below. 2HCl

Zigmanuir [339]

The first step is to find the number of moles of OH⁻ that reacted with the HCl. To do this multiply 2.00L by 1.50M to get 3 moles of Ca(OH)₂. Then you multiply 3 by 2 (there are 2 moles of OH⁻ per every 1 mole of Ca(OH)₂) to get 6 moles of OH⁻. That means that you needed 6 moles of HCl since 1 mole of HCl contains 1 mole of H⁺ and equal amounts H⁺ and OH⁻ reacted with each other. To find the molarity of the HCl solution you need to divide 6mol by 1L to get 6M. Tat means that the concentration of the acid was 6M.

I hope this helps. Let me know if anything was unclear.

5 0

3 years ago

Read 2 more answers

When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t

satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

$Q=C\times Delta T$

$Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J$

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = $\frac{15.3 g}{85 g/mol}=0.18 mol$

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

$\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ$

1 J = 0.001 kJ

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

8 0

3 years ago