Which kingdom does extreme microbe belong to

What does the lysosome do

Ostrovityanka [42]

Answer: They digest excess or worn out organelles, food particles, and engulfed viruses or bacteria.

Explanation:

6 0

4 years ago

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If an equal number of moles of the weak acid HF and the strong base KOH are added to water, is the resulting solution acidic, ba

s344n2d4d5 [400]

Answer:

Being a weak acid and a strong base, where it is diluted in a neutral medium such as water, the basic medium predominates, almost alkaline pH.

Explanation:

The acidity of the solution, being weak, means that its pH is not so low, therefore it will be easier to reach the values of 7 or 7 where alkalinity or basity is indicated.

6 0

3 years ago

Explain the advantage of using scientific names,

posledela

Answer:

Clarity and precision - these names are unique with each creature having only one scientific name. Helps avoid confusion created by common names. 3. Universal recognition - scientific names are standardised and accepted universally

Explanation:

4 0

3 years ago

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$

$\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}$

The above expression is the expression for equilibrium constant of the total equation, which is:

$2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c$

Hence, the equilibrium constant for the total reaction is $4.09\times 10^{-6}$

8 0

3 years ago

Give an example of an element that would be expected to share similar chemical properties with beryllium, calcium, and strontium

Aleks04 [339]

Strontium possibly but I’m not 100% sure

4 0

4 years ago