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4 years ago

1) A toy car with a mass of 2.35 kg is pushed with a force of 21.6 N. What is the acceleration of the car?

Chemistry

1 answer:

Nataly_w [17]4 years ago

6 0

Answer:

The acceleration of the car is 9,19 m/s2

Explanation:

We use the formula: F=m x a---> a=F/m

a=21,6N/ 2,35kg 1N is 1kgxm/s2

a=21,6 kg x m/s2 x 2,35 kg

a=9,191489362 m/s2

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In the formation of potassium chloride, the product, KCl, has less energy than the reactants, K and Cl2. This reaction is .

k0ka [10]

if the product has lesser energy than the reactants it means enthalpy change value is positive.so it is endothermic reaction where the heat is absorbed.so kcl will formation will have positive enthalpy change .

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3 years ago

Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic ac

Elza [17]

Answer:

$H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1$

$H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2$

$HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3$

Explanation:

Hello!

In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:

$H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1$

$H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2$

$HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3$

Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:

Ka1 > Ka2 > Ka3

Best regards!

3 0

3 years ago

A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t

pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = $pK_a=9.31$

First we have to calculate the moles of HCN and NaCN.

$\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole$

and,

$\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole$

The balanced chemical reaction is:

$HCN+NaOH\rightarrow NaCN+H_2O$

Initial moles 0.1116 0.0461 0.08978

At eqm. (0.1116-0.0461) 0 (0.08978+0.0461)

0.0655 0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=9.31+\log (\frac{0.1359}{0.0655})$

$pH=9.63$

Therefore, the pH of the solution is, 9.63

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3 years ago

Given the following unbalanced equation:

Lorico [155]

<h3>Answer:</h3>

8.01 mol MgO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles
Compounds

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + O₂ → MgO

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 8.01 moles Mg

[Solve] moles MgO

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Mg → 2 mol MgO

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 8.01 \ mol \ Mg(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 8.01 \ mol \ MgO$

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3 years ago

. Select the correct statement below a) There are two sp2 hybrid orbitals. b) There is one sp hybrid orbital c) There are four s

photoshop1234 [79]

Answer:

C. There are four sp3 hybrid orbital.

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3 years ago