All categories

3 years ago

An acid-base reaction will always produce water and which one of the

Chemistry

1 answer:

Nonamiya [84]3 years ago

8 0

Answer: Option D) Salt

Explanation:

An example of acid-base reaction is

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

Acid + base → Salt + Water.

Here, water H2O and salt, NaCl are produced.

Usually, the acid, aqeous HCl reacts completely with an appropriate amount of the base, aqueous NaOH to produce salt, aqueous NaCl and water, liquid H2O only, in a process also known as neutralization reaction

You might be interested in

1. A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equa

kvv77 [185]

Answer:

1B +4St+1Y+3lc——-> BSt4Ylc3

Explanation:

I only know the answer for the first question.

4 0

3 years ago

The decay of which radioisotope can be used to estimate the age of the fossilized remains of an insect?

Dmitriy789 [7]

(4) C-14 Carbon is found in all living organisms.

3 0

3 years ago

Read 2 more answers

How many nanometers are in 0.0006245101 km?

maw [93]

Answer:

624510100

Explanation:

Doing a conversion factor:

$0,0006245101[km]*\frac{1000[m]}{1 km} *\frac{1x10^{9} nanometer}{1 m} =624510100 [nanometer]$

5 0

3 years ago

In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec (1 min = 60 sec)

Now put all the given values in this formula, we get

$Q=10A\times 600s=6000C$

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = $2\times (1.6\times 10^{-19})$

Number of atoms deposited = $\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}$ atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = $\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113$ moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g

Mass of Copper Deposited = $63.5\times 0.03113 =1.98g$

Therefore, the mass of copper deposit is, 1.98 grams

5 0

3 years ago

Given the following information:Br2 bond energy = 193 kJ/mol F2 bond energy = 154 kJ/mol 1/2Br2(g) + 3/2F2(g) -> BrF3(g) = –3

arlik [135]

Answer:

C) 712 KJ/mol

Explanation:

ΔH°r = Σ Eb broken - Σ Eb formed

1/2Br2(g) + 3/2F2(g) → BrF3(g)

∴ ΔH°r = - 384 KJ/mol

∴ Br2 Eb = 193 KJ/mol

∴ F2 Eb = 154 KJ/mol

⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)

⇒ Σ Eb broken = (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol

∴ Eb formed: Br-F

⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r

⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )

⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol

8 0

3 years ago