Is " Energy created " an Endothermic reaction or an Exothermic reaction?

Ribosome are found ____

evablogger [386]

It's found inside of a cell attached to the cytoplasm and attached to the endoplasmic reticulum. hope this helps

6 0

2 years ago

State the law of conservation of mass

Katen [24]

It’s basically that’s any system that’s closed to all transfers of matter and energy the mass of the system has to remain constant over time because they can’t change meaning you can’t add or remove from it

8 0

2 years ago

Water has a heat of fusion of about 300J/gram and a heat capacity for liquid water of a about 4J/gC. How much heat is needed to

amm1812

Answer:

The total heat required is 3.4 kJ

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. So, the amount of heat a body receives or transmits is determined by:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case you know;

Q= ?

c= 4 $\frac{J}{g*C}$

m= 10 g
ΔT= Tfinal - Tinitial= 10 C - 0 C= 10 C

Replacing:

$Q= 4\frac{J}{g*C} *10 g * 10 C$

Solving:

Q1= 400 J

On the other hand, you must determine the heat required to convert 0 ∘ C of ice to 0 ∘ C of liquid water by:

Q2=m*heat of fusion

Q2=10 g* 300 $\frac{J}{g}$

Q2= 3,000 J

The total heat required is:

Q= Q1 + Q2= 400 J + 3,000 J

Q= 3,400 J= 3.4 kJ (1 kJ= 1,000 J)

The total heat required is 3.4 kJ

5 0

3 years ago

Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1

bekas [8.4K]

Explanation:

$2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)$

Equilibrium constant of reaction = $K=154$

Concentration of NO = $[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M$

Concentration of bromine gas = $[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M$

Concentration of NOBr gas = $[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M$

The reaction quotient is given as:

$Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}$

$Q=328.06$

$Q>K$

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced. False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0

2 years ago

2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)

ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

$n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}$

2. Calculate the mass of KCl

$m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}$

3. Prepare the solution

Measure out 224 g of KCl.
Dissolve the KCl in a few hundred millilitres of distilled water.
Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.

8 0

2 years ago