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zlopas [31]
3 years ago
8

The equation for the chemical reaction shown is not balanced. What number should replace the question mark to balance this equat

ion?
2Ag2O → ?Ag + O2









A.
2







B.
3







C.
1







D.
4
Chemistry
1 answer:
Anton [14]3 years ago
3 0
It will be D.4 because you have two lots of 2Ag atoms so its 4 in total
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Greater than ___________ mcg/dl is considered a high level of morning cortisol.
Natalija [7]

Greater than 23mcg/dl is considered a high level of morning cortisol.

Normally, cortisol levels rise during the early morning hours and are highest about 7AM. They drop very low in the evening and during the early phase of sleep. If you do not have this daily change (diurnal rhythm) in cortisol levels, you may have overactive adrenal glands. This condition is called Cushing's syndrome.

5 0
3 years ago
how many ml of a 22.5% (v/v) ethanol solution would you need to measure out in order to have 12.5 ml of ethanol ?
Aleks [24]
Volume percent<span> or </span>volume/volume percent<span> (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:

12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed

Hope this answers the question.</span>
8 0
4 years ago
What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
What is the mass of 1.7 × 1023 atoms of zinc
fiasKO [112]

We are given with a compound, Zinc (Zn) having a 1.7 x 10 ^23 atoms. We are tasked to solve for it's corresponding mass in g. We need to find first the molecular weight of Zinc, that is

Zn= 65.38 g/mol

Not that 1 mol=6.022x10^{23} atoms, hence,

1.7 x 10 ^23 atoms x 1 mol/6.022x10^{23} atoms x65.38 g/ 1mol

=18.456 g of Zn

 

Therefore, the mass of Zinc 18.456 g

3 0
3 years ago
A 1.00 g sample of n-hexane (C6H14) undergoes complete combustion with excess O2 in a bomb calorimeter. The temperature of the 1
Nadya [2.5K]

Answer:

-5,921x10⁶J/mol

Explanation:

Internal energy change (ΔU) for the reaction of combustion in the bomb calorimeter is:

ΔU = q calorimeter + q solution

Where:

q calorimeter is Ccal×ΔT (Ccal=4042J/°C) and (ΔT is 29,30°C-22,64°C=<em>6,66°C</em>)

q solution is c×m×ΔT (c= 4.184 J/g°C), (m=1502g H₂O), (ΔT is 29,30°C-22,64°C=<em>6,66°C</em>)

Replacing:

ΔU = 26920J + 41854J = <em>68774 J</em>

This energy is per g of n-hexane, now, per mole of n-hexane:

\frac{68774J}{1gHexane} *\frac{86,1g}{1mol}= -<em>5,921x10⁶J/mol</em>

<em>-negative because the energy is produced-</em>

I hope it helps!

5 0
4 years ago
Read 2 more answers
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