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monitta
3 years ago
7

74 calories to kilocalories

Chemistry
1 answer:
grigory [225]3 years ago
4 0

Answer:74 calories to kilocalories  =0.074

Hope this helps

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Fill in the blank with the correct answer. The longer the handle the __________ the mechanical advantage and the greater the inc
ASHA 777 [7]

Answer:

Greater

Explanation:

The longer the handle, the greater the mechanical advantage and the greater the increase in force that is applied to the bolt.

Mechanical advantage is the rate at which force is multiplied.

  • It simply compares the output force to the input force.
  • M.A is the force increasing tendency of a tool.
  • The longer  the handle, the more the mechanical advantage and the lesser the applied force.
7 0
3 years ago
What element has an electron<br> configuration of 2-8-8-1:
Afina-wow [57]
Ойлголоо, уучлаарай Ойлголоо, уучлаарай /; coo
4 0
2 years ago
If a clot were made up of a mass of proteins, what change in the proteins led to the formation of a clot?
S_A_V [24]

Answer: The proteins were no longer soluble in the blood.

3 0
3 years ago
What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?
Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

6 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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