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2 years ago

1. what is the pressure in space equal to?

Chemistry

1 answer:

zheka24 [161]2 years ago

6 0

Answer:

1. The pressure in outer space is so low that many consider it as non-existant. It has a pressure of 1.322 × 10−11 Pa. Pressure may be detected from the molecule of air or water hitting you. Since there is very little air and hardly ever water hitting you in space, pressure is almost zero or negligible.

2. Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, 14.70 pounds per square inch, 1,013.25 × 103 dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

3. 0.46J/gC

explanation.The specific heat capacity of a material is given by:

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C_s = \frac{Q}{m \Delta T}C s = mΔTQ

where

Q is the amount of heat absorbedm is the mass

\Delta TΔT is the variation of temperature

For the piece of iron in the problem:

m = 15.75 gm=15.75g

Q=1086.75 JQ=1086.75J

\Delta T=175 C-25 C=150^{\circ}ΔT=175C−25C=150∘

Substituting into the equation,

C_s = \frac{1086.75 J}{(15.75 g)(150^{\circ}C)}=0.46 J/gCC s = (15.75g)(150 ∘ C)1086.75J = 0.46J/gC

4. 207 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of 1 g of a given substance by 1∘C.

#I hope it's help

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What change would most likely decrease the rate of a chemical reaction? A) maintaining a temperature close to room temperature B

Nutka1998 [239]

Answer:

D) cooling of the reaction mixture

Explanation:

Increase in temperature speedens the rate of a chemical reaction

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2 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

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