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2 years ago

14

A white powder, when heated, produces a colourless gas and a black solid. Is the white powder an element? I need reasons please

:)

1 answer:

NeTakaya2 years ago

4 0

Answer:It is not an element because elements are the purest form of a substance; hence, they are no longer broken down by heating

Explanation:

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How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+

Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21 = 13.1 g

3 0

2 years ago

Sodium thiosulfate (Na2S2O3), photographer’s

vova2212 [387]

3Na2S2O3 + AgBr ------>Na5[Ag(S2O3) 3] +NaBr

from equation 3 mol 1 mol

given x mol 0.10 mol

x= (3*0.10)/1=0.30 mol Na2S2O3

Answer: 0.30 mol Na2S2O3

3 0

2 years ago

Read 2 more answers

The combustion of propane is best described as an

GalinKa [24]

<h3><u>Answer;</u></h3>

Exothermic chemical reaction

<h3><u>Explanation;</u></h3>

Exothermic reactions are chemical reactions where the products have less energy that the reactants.
Exothermic reactions give off energy, usually in the form of heat, while endothermic reactions absorb energy.
The combustion of propane is definitely an exothermic reaction because it generates a lot of heat.

5 0

2 years ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>( $K_{a}$ )<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

<u>The acid dissociation constant: </u> $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0

2 years ago

Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

$\frac{r_{cation}}{r_{anion}}$

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, $Na^+$ = 102 pm

Radius of cation, $Br^-$ = 196 pm

$\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520$

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0

2 years ago