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Artyom0805 [142]

3 years ago

6

A sample of chlorine gas is stored at a temperature of -25oC. If the chlorine gas's temperature is raised to 52oC, its volume in

creases to 2.60 L. What is the initial volume of the gas?

1 answer:

aalyn [17]3 years ago

4 0

Answer:

whats ur answer options

Explanation:

...

its b

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The inhibition caused by the final end product of a reaction is called

Troyanec [42]

Answer:

Non competitive inhibition

Explanation:

Hello,

During enzymatic catalysis, the active sites could be occupied by the very same products' molecules turning out into an inhibition (the reaction starts to slow down since to active places are available for the reagents to react). Nonetheless this inhibition is not competitive as long as the product does not react due to the active sites it is occupying.

Best regards.

5 0

3 years ago

A 25.0g tin sample was placed in boiling water at 99.5 C until it had the same temperature. Afterwards, the tin sample was place

MrMuchimi

Answer : The specific heat of tin is, 0.213 J/g.K

Explanation :

Formula used :

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = amount of heat lost = -399.4 J

c = specific heat capacity of tin = ?

m = mass of tin = 25.0 g

$T_{final}$ = final temperature = $24.5^oC=273+24.5=297.5K$

$T_{initial}$ = initial temperature = $99.5^oC=273+99.5=372.5K$

Now put all the given values in the above formula, we get:

$-399.4J=25.0g\times c\times (297.5-372.5)K$

$c=0.213J/g.K$

Therefore, the specific heat of tin is, 0.213 J/g.K

7 0

3 years ago

5. Which pair of elements would combine to forin an ionic compound?

Natasha_Volkova [10]

C.silicon and oxygen is the answer

5 0

3 years ago

When measuring the volume of an irregularly shaped object you would measure the amount of water____ or pushed away

Over [174]

Displacement is the answer

3 0

3 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago