You run a 5k for charity. How many feet do you run?

What are some double replacement reactions you see every day

Cerrena [4.2K]

Answer:

Sodium Bicarbonate (β-3) + Vinegar
Lead Nitrate + Potassium iodide

Explanation:

Baking Soda and vinegar cause an explosion, in which the bicarbonate and vinegar are replaced by nitrate (∨) and oxide (Ф.) When you combine lead nitrate (Δω) with potassium iron, you also see the ingredients you combined disappear, which shall cause a replace reaction

6 0

2 years ago

3.0 cm x 4.0 cm x 1.0 cm [?]cm^3

CaHeK987 [17]

Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

=3cm × 4 cm × 1cm

= 12cm^2×1cm (4×3=12)

= 12cm^3 (12×1=12)

Therefore, theanswer is 12 cm^3.

Hope it helps..

4 0

3 years ago

Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s

NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

$n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3$

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0

2 years ago

Calcium ions are removed from the sarcoplasm of the cardiac cell by the action of the sarcoplasmic reticulum (SR) Ca2+-ATPase pu

Ivan

Answer:

Sodium-calcium exchanger on the sarcolemma

Explanation:

The sodium-calcium exchanger which is an antiporter membrane protein, removes calcium from cells. The energy of the electrochemical gradient of sodium is used by allowing it to flow down its gradient while moving across the plasma membrane

7 0

3 years ago

g Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at

Alborosie

Answer : The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})$

where,

$\Delta G^o$ = standard Gibbs free energy = -14.1 kJ/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

Q = reaction quotient

$[A]_{inside}$ = concentration inside the cell

$[A]_{outside}$ = concentration outside the cell

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

$-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})$

$\frac{[A]_{inside}}{[A]_{outside}}=296.2$

Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

3 0

2 years ago