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Bess [88]
3 years ago
11

The atomic number of sodium (Na) is 11. This is also the number of which of the following?

Chemistry
1 answer:
attashe74 [19]3 years ago
3 0
D. Protons, Atomic number is the number of protons in an atom.
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Write a net ionic equation to show that hydroiodic acid, hi, behaves as an acid in water.
OlgaM077 [116]

Answer:

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Explanation:

The HI donates a proton to the water, converting it to a hydronium ion

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Thus, the HI is behaving like a Brønsted acid.

3 0
3 years ago
Read 2 more answers
GIVING 20 POINTS!! Apply Concepts Two chemists each plan an investigation in which they will combine hydrogen gas with chlorine
Montano1993 [528]

Answer:

Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2

Explanation:

1. The law of multiple proportions says that when elements form compounds, the proportions of the elements in those chemical compounds can be expressed in small whole number ratios.  This means that regardless of whether 1000 times more of the products are used, the reactivity of the products is established by the chemical reaction

2. The law of multiple proportions is an extension of the law of definite composition, which states that compounds will consist of defined ratios of elements.

3. A reaction with more reactants will need more care because more products are produced, which can be toxic

4. H2 and Cl2 reactivity does not depend on the quantities but the chemical properties of each compound

5 0
3 years ago
El oxígeno y el azufre reaccionan con el cobre para formar óxido de cobre (CuO) y sulfuro de cobre (CuS), respectivamente. ¿Qué
oksian1 [2.3K]
Yes it need to be like that cause when it like that it like that
5 0
3 years ago
How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca
leva [86]

Answer:

The answer to your question is: 6.8 g of water

Explanation:

Data

2.6 moles of HCl

1.4 moles of Ca(OH)2

                           2HCl     +     Ca(OH)2    →        2H2O    +      CaCl2

MW                   2(36.5)               74                       36 g               111 g

                          73g                

                            1 mol of HCl ----------------  36.5 g

                           2.6 mol           --------------    x

                              x = (2.6 x 36.5) / 1   = 94.9 g

                           1 mol of Ca(OH)2 --------------   74 g

                         1.4 mol                  ---------------   x

                            x = (1.4 x 74) / 1  = 103.6 g

Grams of water

                        73 g of HCl ------------------   36g of H2O

                        94.9 g        -------------------    x

                     x = (94.9 x 36) / 73 = 46.8 g of water

6 0
3 years ago
When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?
irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

                 AgNO₃(aq)  +  NH₄Cl(aq)   ⇒   AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of     AgNO₃(aq)  to  NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

                     1 mol of NH₄Cl  ---------------- 1 mol of AgCl

              0.00188 mol of NH₄Cl ------------- x

                     x = (0.00188 x 1) /1

                     x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

                         143.5 grams of AgCl -------------- 1 mol

                         x -------------------------------------------0.00188 moles of AgCl

                          x = (0.00188 x 143.5) / 1

                          x = 0.269 grams of AgCl

8 0
3 years ago
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