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3 years ago

In the number 48.93, which digit is estimated?

Chemistry

1 answer:

Fudgin [204]3 years ago

7 0

The answer is 3.

Explanation:
It’s the last number and it can’t be 9 because then it would be 48.9 and no 3.

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2CO2 + 3H2O + 4Br2 C = O = H = Br =

devlian [24]

Explanation:

2CO2 + 3H2O + 4Br2

C = 2

O = 5

H = 6

Br =8

8 0

3 years ago

Need help ASAP !!

Kitty [74]

Answer:

it's a segment

Explanation:

it has multiple end points

6 0

3 years ago

When the temperature of a gas changes, it's volume decreases from 12 cm3 to 7 cm3 if the final temperature is measured to be 18°

ElenaW [278]

Answer:

The initial temperature is 499 K

Explanation:

Step 1: Data given

initial volume = 12 cm3 = 12 mL

Final volume = 7 cm3 = 7mL

The final temperature = 18 °C = 291 K

Step 2: Calculate the initial temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.012 L

⇒with T1 = the initial volume = ?

⇒with V2 = the final volume 0.007 L

⇒with T2 = The final temperature = 291 K

0.012 / T1 = 0.007 / 291

0.012/T1 = 2.4055*10^-5

T1 = 0.012/2.4055*10^-5

T1 = 499 K

The initial temperature is 499 K

8 0

3 years ago

Indicate whether the following balanced equations involve oxidation-reduction. Check all that apply. Check all that apply. 2H2SO

guapka [62]

Answer : The balanced equations involve oxidation-reduction are:

(a) $2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)$

(b) $3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)$

(c) $NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is:

$2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)$

This reaction involve oxidation-reduction reaction because the oxidation state bromine changes from (-1) to (0) which shows oxidation and sulfur changes from (+6) to (+4) which shows reduction.

(b) The given chemical reaction is:

$3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)$

This reaction involve oxidation-reduction reaction because the oxidation state sulfur changes from (+4) to (+6) which shows oxidation and nitrogen changes from (+5) to (+2) which shows reduction.

(c) The given chemical reaction is:

$NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)$

This reaction involve oxidation-reduction reaction because the oxidation state iodine changes from (-1) to (+5) which shows oxidation and chlorine changes from (+5) to (-1) which shows reduction.

(d) The given chemical reaction is:

$PBr_3(l)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HBr(aq)$

This reaction does not involve oxidation-reduction reaction because the oxidation state of element present on reactant and product side are same.

3 0

3 years ago

According to Le Châtelier's principle, how will an increase in concentration of a reactant affect the equilibrium system?

Svetach [21]

Answer:

Shift it toward the products

Explanation:

Imagine this equilibrium:

2A + B ⇄ A₂B

Let's think the expression for Kc:

Kc = [A₂B] / [A]² . [B]

Now, we have this situation:

2A ↑ + B ⇄ A₂B

As we are not in equilibrium, we have to think that Kc would decrease.

↓Kc = [A₂B] / [A]² ↑ . [B]

so we have to use Qc (reaction quotient). For this case

Qc < Kc, so the reaction → Product side

We have to make more product to counteract the reagent increase and thus return the system to equilibriumr

3 0

3 years ago

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