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3 years ago

Help I really need help and don't search it on go.ogle this is a school assignment and i can get in trouble thank you sm :)

Chemistry

1 answer:

kupik [55]3 years ago

4 0

Answer:

Nuclear fission is almost 5 times times more efficient than traditional fossil fuels at producing energy. That's a lot of energy packed into a small space. Nuclear energy is more efficient, which means it uses less fuel to power the plant and produces less waste.

Explanation:

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Data for the decomposition of hydrogen peroxide at some set temperature T is provided below. The rate law depends only on the co

Viktor [21]

Answer:

$0.00442 s^{-1}$ is the value of the rate constant.

Explanation:

$2 H_2O_2\rightarrow 2 H_2O + O_2$

Let the order of the reaction be x.

The rate law of the reaction can be written as:

$R=k[H_2O_2]^x$

1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds to 60 seconds.

$R=-\frac{0.697 M-0.882 M}{60 s-0 s}=0.00308 M/s$

$0.00308 M/s=k[0.697 M]^x$ ..[1]

2. Rate of the reaction when concentration changes from 0.697 M to 0.566 M in 240 seconds to 360 seconds.

$R=-\frac{0.236M-0.372M}{120s-60 s}=0.00227 M/s$

$0.00218 M/s=k[0.236 M]^x$ ..[2]

[1] ÷ [2]

$\frac{0.00308 M/s}{0.00227 M/s}=\frac{k[0.697 M]^x}{k[0.236M]^x}$

Solving fro x:

x = 0.92 ≈ 1

$R=k[H_2O_2]^1$

$0.00308 M/s=k[0.697 M]^1$

$k=\frac{0.00308 M/s}{[0.697 M]^1}=0.00442 s^{-1}$

$0.00442 s^{-1}$ is the value of the rate constant.

5 0

4 years ago

Use the following balanced reaction to solve:

Naily [24]

Answer: 60.7 g of $PH_3$ will be formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}$

$\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles$

The balanced chemical reaction is

$P_4(s)+6H_2(g)\rightarrow 4PH_3(g)$

$H_2$ is the limiting reagent as it limits the formation of product and $P_4$ is the excess reagent.

According to stoichiometry :

6 moles of $H_2$ produce = 4 moles of $PH_3$

Thus 2.68 moles of $H_2$ will produce= $\frac{4}{6}\times 2.68=1.79moles$ of $PH_3$

Mass of $PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g$

Thus 60.7 g of $PH_3$ will be formed by reactiong 60 L of hydrogen gas with an excess of $P_4$

3 0

3 years ago

What is the formula of titanium(IV) bromide?

Anna [14]

Answer:

$\boxed{\text{SnBr}_{4}}$

Explanation:

The name tells you that this is a binary compound (contains two elements).

It contains a metal and a nonmetal, so it is a binary ionic compound. The general rule is:

Name of compound = name of metal name of ion (two words)

Name of metal = tin(IV), so the tin ion has a charge of 4+

Name of ion = bromide. Br is in Group 17, so bromide ion has charge of 1-.

$\rm Sn$^{4+}$ + 4Br$^{-} \longrightarrow \,$ SnBr$_{4}$\\\\\text{The formula is} \boxed{\textbf{SnBr}_{4}}$

3 0

4 years ago

Which of the following materials can conduct electricity?

Lorico [155]

Answer:

plastics, Styrofoam, paper, rubber, glass and dry air.

Explanation:

7 0

3 years ago

Read 2 more answers

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of t

Dmitry_Shevchenko [17]

Answer:

When we add 0.33L of KOH only the HCl solution will be neutralized. When we add 1.4 L of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.

Explanation:

Step 1: Data given

The solution has 0.100 M HCl and 0.210 M H2SO4

Molarity KOH = 0.150 M

Volume of acid solution = 500 mL = 0.5 L

Step 2: Calculate moles of HCl

Moles HCl = Molarity HCl * volume

Moles HCl = 0.100 M * 0.5 L

Moles HCl = 0.05 moles

Step 3: Calculate moles of H2SO4

Moles H2SO4 = 0.210 M * 0.5 L

Moles H2SO4 = 0.105 moles

Step 4: The balanced neutralization reaction of KOH with HCl can be written as:

KOH + HCl → KCl + H2O

The mole ratio is 1:1

This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH

Step 5: The balanced neutralization reaction of KOH with H2SO4 can be written as:

2KOH + H2SO4 → K2SO4 + 2H2O

The mole ratio KOH: H2SO4 is 2:1

This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH

Step 6: Calculate volume of KOH needed to neutralize the solution

To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed

To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed

When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.

4 0

3 years ago