Which tiny, round “factory” puts together protein and is often found in the endoplasmic reticulum?

What about 50 g of water? I need help what this

ELEN [110]

Answer:

3.38 Tablespoons

10.14 Teaspoons

0.21 U.S. Cups

0.18 Imperial Cups

0.20 Metric Cups

50.00 Milliliters

Explanation:

3 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

What is the reaction energy q of this reaction? use c2=931.5mev/u. express your answer in millions of electron volts to three si

Ann [662]

₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
Mass defect Δ m = 236.052595 - 235.854000 = 0.198 amu
Reaction energy released Q = Δ m * 931.5
= 0.198 * 931.5 = 185 MeV

5 0

2 years ago

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the pressure is increased? There will not be a shift in

Tamiku [17]

When the pressure is increased, the equilibrium will shift to the left to offset the pressure increase. Equilibrium shifting to the left side is favored because the left side has fewer moles of gas than the number of moles of gas on the right-hand side and because it exerts less pressure. Therefore, the answer is there will be a shift toward the reactants.

6 0

3 years ago

Read 2 more answers

Help please stuck and confused

Alecsey [184]

I’m pretty sure it’s none of the above cause I’m googling it a bunch and and it say you use the dideoxy method

7 0

2 years ago