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RSB [31]
3 years ago
11

What are the products of this chemical equation? HCl(aq) + NaOH(aq) → NaCl(aq) + H2(g) + O2(g) NaH(aq) + Cl2(aq) + H2(g) + O2(g)

Cl2(g) + H2(g) + O2(g) NaCl(aq) + H2O(l)
Chemistry
2 answers:
Iteru [2.4K]3 years ago
7 0

Answer:

The correct answer is:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Explanation:

When aqueous solution of hydrogen chloride reacts with aqueous solution of sodium hydroxide to give aqueous solution of sodium chloride and water.

HCl(aq) + NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)

1 mole of hydrogen chloride reacts with 1 mol of sodium hydroxide to give 1 mole of sodium chloride and 1  mole of water.

Hatshy [7]3 years ago
5 0
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H₂O(l)

:•)
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1. What would happen to the current model of the atom if new information about its structure is discovered in the future?
Semenov [28]
1-Option A. Current model will be revised. 
2-Option D. Nucleus
3-<span>Option B. </span>Counting the no. of protons.
6 0
3 years ago
10. A small gold nugget has volume of 0.87 cm3. What is its mass if the density of gold is 19.3 g/cm3?
bulgar [2K]

Answer:

16.791 grams

Explanation:

The density formula is:

d=\frac{m}{v}

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

d*v=\frac{m}{v}*v

d*v=m

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

d= 19.3 g/cm^3\\v-0.87 cm^3

Substitute the values into the formula.

m=d*v

m= 19.3 g/cm^3*0.87 cm^3

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

m=19.3 g*0.87

m=16.791 g

The mass of the gold nugget is 16.791 grams.

8 0
3 years ago
I’ll mark brainliest please
DaniilM [7]
I’m assuming your just writing the formula? If so
Potassium chloride: KCL
Potassium nitride: KNO2
Potassium sulfide: K2S
calcium chloride: CaCl2
Calcium nitride: Ca3N2
Calcium sulfide: CaS
Silver chloride: AgCl
Silver nitride: Ag3N
Silver sulfide: Ag2S
Manganese (||) chloride: MnCl2
Manganese (||) nitride: Mn3N2
Manganese (||) sulfide: MnS
5 0
2 years ago
A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10
Anton [14]

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

<h3>What is the Henderson-Hasselbalch equation?</h3>

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

= 4.75

Therefore, 4.75 is the pH of the buffer.

Learn more about the Henderson-Hasselbalch equation here:

brainly.com/question/27751586

#SPJ4

6 0
1 year ago
Help me please i put nine points!
Ahat [919]

Answer:

eons.

It would be D. An eon.

3 0
2 years ago
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