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2 years ago

Which two chemical equations model double-replacement reactions?

Chemistry

2 answers:

lora16 [44]2 years ago

6 0

Answer:

C. AgNO3 + LICI - AgCl + LINO3

Explanation:

Here the chemical equation for double displacement reaction is

AgNO3 + LICI - AgCl + LINO3

The reaction in which two compounds react together to form two other compounds by mutual exchange of their ions is called double displacement reaction.

Hope it will help :)

ozzi2 years ago

4 0

Answer:

A and C just got it right ;)

Explanation:

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What equation is used to calculate the pH of a solution?

Galina-37 [17]

Answer:

pH = -log[H3O+] or pH = -log[H+]

Explanation:

pH is the negative log of hydronium concentration or hydrogen ion concentration. The two are essentially interchangeable.

7 0

3 years ago

How many 0 atoms are in zn(CIO3)2

Andrew [12]

Answer:

Six atoms.

Explanation:

The number outside the brackets (2), should be multiplied by the number of atoms inside the brackets. Thus, there will be 2 chlorine atoms and 6 oxygen atoms.

6 0

3 years ago

Read 2 more answers

Calculate the mass percent for all components in a solution containing the following. 0.350 Kg of water, 5.4 moles of ammonia an

Alina [70]

Answer:

% Water = 54.96%

% Ammonia = 0.14%

% Cobalt (II) Nitrate = 30.62%

Explanation:

To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:

Mass Water ⇒ 0.350 kg Water = 350 g water

Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol = 91.8 g

Mass cobalt (II) nitrate ⇒ 195.0 g

Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate

Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g

To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:

% Water ⇒ 350/636.8 * 100% = 54.96%

% Ammonia ⇒ 91.8/636.8 * 100% = 0.14%

% Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%

8 0

3 years ago

QUESTION THREE

BartSMP [9]

Answer:

Odds to be given for an event that either Romance or Downhill wins is 11:4

Explanation:

Given an odd, r = a : b. The probability of the odd, r can be determined by;

Pr(r) = $\frac{a}{b}$ ÷ (

So that;

Odd that Romance will win = 2:3

Pr(R) = $\frac{2}{3}$ ÷ (

= $\frac{2}{3}$ ÷ $\frac{5}{3}$

= $\frac{2}{5}$

Odd that Downhill will win = 1:2

Pr(D) = $\frac{1}{2}$ ÷ (

= $\frac{1}{2}$ ÷ $\frac{3}{2}$

= $\frac{1}{3}$

The probability that either Romance or Downhill will win is;

Pr(R) + Pr(D) = $\frac{2}{5}$ + $\frac{1}{3}$

= $\frac{11}{15}$

The probability that neither Romance nor Downhill will win is;

Pr(neither R nor D) = (1 - $\frac{11}{15}$ )

= $\frac{4}{15}$

The odds to be given for an event that either Romance or Downhill wins can be determined by;

= Pr(Pr(R) + Pr(D)) ÷ Pr(neither R nor D)

= $\frac{11}{15}$ ÷ $\frac{4}{15}$

= $\frac{11}{4}$

Therefore, odds to be given for an event that either Romance or Downhill wins is 11:4

8 0

3 years ago

Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze

Rashid [163]

Answer:

a. Δ $H^0_{rxn} = -108.0\frac{kJ}{mol}$

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate Δ $H^0_{rxn}$ , for the gas-phase reaction ) using the Hess's Law.

Δ $H^0_{rxn}$ = $E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}$

Given from the question, the table below shows the corresponding Δ $H^0_{t}(kJ/mol)$ for each compound.

Compound $H^0_{t}(kJ/mol)$

Liquid EO -77.4

$CH_4_(g_)$ -74.9

$CO_(g_)$ -110.5

If we incorporate our data into the above previous equation; we have:

Δ $H^0_{rxn}$ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

= $-108.0 \frac{kJ}{mol}$

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

$T_{initial}$ = 93.0°C &

the enthalpy of vaporization (Δ $H^0_{vap}$ ) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as = $\frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}$