A measurement must include both number and a ?

__ SnCl4 → __ Sn + __ Cl2

Kisachek [45]

Answer:

Sn (s) + 2 Cl2 (g) → SnCl4 (l)

This is an oxidation-reduction (redox) reaction:

4 Cl0 + 4 e- → 4 Cl-I

(reduction)

Sn0 - 4 e- → SnIV

(oxidation)

Cl2 is an oxidizing agent, Sn is a reducing agent.

Reactants:

Sn

Names: Tin source: wikidata, accessed: 2019-09-07source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Sn source: wikidata, accessed: 2019-09-07, Element 50 source: wikidata, accessed: 2019-09-07

Appearance: White crystalline powder source: ICSC, accessed: 2019-09-04; Gray to almost silver-white, ductile, malleable, lustrous solid. source: NIOSH NPG, accessed: 2019-09-02

Cl2

Names: Chlorine source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Molecular chlorine source: NIOSH NPG, accessed: 2019-09-02

Appearance: Greenish-yellow compressed liquefied gas with pungent odour source: ICSC, accessed: 2019-09-04; Greenish-yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.] source: NIOSH NPG, accessed: 2019-09-02

Products:

SnCl4 – Tetrachlorostannane source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02, Tin tetrachloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(IV) chloride source: wikipedia, accessed: 2019-09-28

Other names: Stannic chloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(iv) chloride (anhydrous) source: ICSC, accessed: 2019-09-04

Appearance: Colorless to slightly yellow fuming liquid source: wikipedia, accessed: 2019-09-28; Colourless or slightly yellow fuming liquid with pungent odour source: ICSC, accessed: 2019-09-04

6 0

2 years ago

Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come

Elden [556K]

Answer:

B

Explanation:

For solving this we need a heat balance

$Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}$

By changing the corresponding relations, we have

$m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\$

By cancelling similar factor, we obtain

$\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\$

Which means that the change of temperature in A is twice the change of B

3 0

3 years ago

What is the mass of a rectangular cube with dimensions of 3 cm x 2 cm x to 10 cm and a density of 10 g/cm^3

marta [7]

d=m/V so m=Vd

m=60cm^3 * 10g/cm^3=600g

3 0

3 years ago

Read 2 more answers

Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan

Valentin [98]

Answer : The equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

Explanation :

The given main chemical reaction is:

$N_2(g)+2O_2(g)\rightarrow N_2O_4(g)$ ; $K$

The intermediate reactions are:

(1) $N_2O_4(g)\rightarrow 2NO_2(g)$ ; $K_a$

(2) $\frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g)$ ; $K_b$

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) $2NO_2(g)\rightarrow N_2O_4(g)$ ; $\frac{1}{K_a}$

(2) $N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$ ; $(K_b)^2$

Thus, the equilibrium constant for this reaction will be:

$K=\frac{1}{K_a}\times (K_b)^2$

$K=\frac{(K_b)^2}{K_a}$

Thus, the equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

5 0

3 years ago

How would the calculated concentration of the hci be affected if the sodium hydroxide were poured from ma beaker that contained

Mekhanik [1.2K]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the beaker that contained some water before NaOH were added means that the resulting solution in that beaker will be more dilute. When this diluted sodium hydroxide solution is added to HCl (not hci), the reaction below occurs

HCl + NaOH ⇒ NaCl + H₂O

The reaction above is a neutralization reaction. <u>The concentration of the acid (HCl) will reduce when a base (sodium hydroxide) is added and will also reduce more because of the presence of more water (in the base) which normally reduces the concentration of ions present in an acid or a base to become more dilute.</u>

5 0

3 years ago