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3 years ago

Help me please :3 :3 :3 ;3

Chemistry

2 answers:

tester [92]3 years ago

7 0

I think the correct answer is b

bekas [8.4K]3 years ago

6 0

I think the correct answer would be B) Suggestion 1

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Did you write these compounds? KBr 2 NH₂OH 2HNO3

Orlov [11]

The missing components of the neutralisation reaction include the following:

KBr
KBr 2NH4OH
KBr 2NH4OH2HNO2

<h3>What is neutralisation reaction?</h3>

Neutralisation reaction is defined as the type of reaction that leads to the formation of salt and water when an acid and a base reacts.

From the reactions given the missing components are replaced as follows:

HBr + KOH --> KBr + H2O

H2SO4 + 2NH4OH --> (NH4)2SO4 + 2H2O

2HNO3 + Mg(OH)2 --> Mg(NO3)2 + 2H2O

Learn more about acids here:

brainly.com/question/26353151

#SPJ1

7 0

2 years ago

This boggled my brain plzzz help❤

Andrew [12]

A is combustion

B is Combination

C is double replacement

D is decomposition

E is combination

3 0

4 years ago

What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write

LuckyWell [14K]

The balanced chemical equation for reaction of $AgNO_{3}$ and $Na_{2}SO_{4}$ is as follows:

$2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}$

From the balanced chemical equation, 2 mol of $AgNO_{3}$ reacts with 1 mol of $NaNO_{3}$ .

First calculating number of moles of $NaNO_{3}$ as follows:

$M=\frac{n}{V}$

On rearranging,

$n=M\times V$

Here, M is molarity and V is volume. The molarity of $NaNO_{3}$ is given 0.274 M or mol/L and volume 155 mL, putting the values,

$n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol$

Since, 1 mol of $NaNO_{3}$ reacts with 2 mol of $AgNO_{3}$ thus, number of moles of $AgNO_{3}$ will be $2\times 0.04247 mol=0.08494 mol$ .

Now, molarity of $AgNO_{3}$ is given 0.305 M or mol/L thus, volume can be calculated as follows:

$V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL$

Therefore, volume of $AgNO_{3}$ is 278.5 mL.

4 0

4 years ago

Compared to the charge of a proton, the charge of an electron has?

PolarNik [594]

The charge of an electron has a negative charge (-), while protons have a postivie (+) charge.

4 0

3 years ago

slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.

pishuonlain [190]

Answer:

For A: The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

Explanation:

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

For A: Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

6 0

3 years ago