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2 years ago

A van moves with a constant speed

Chemistry

1 answer:

Kitty [74]2 years ago

4 0

Answer:

1.5 hours or 90 minutes

Explanation:

Velocity = d/t

V = 64 mi / hour or 96 mi / 9 min OR 16 mi / 15 min

D = 96 miles

Velocity * Time = Distance

Time = Distance / Velocity

T = 96 mi / 64 mi / hour

96 / 64 = 1.5 mi per hour

T = 1.5 hours or 90 minutes

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Alinara [238K]

Answer:

Use the formula q = m·ΔHv in which q = heat energy, m = mass, and ΔHv = heat of vaporization.

Explanation:

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3 years ago

Hemoglobin (Hb) can be viewed as having two quaternary states, a low oxygen affinity state (T), and a high oxygen affinity state

natta225 [31]

<h3><u>Answer;</u></h3>

The statements that are True are;

Upon binding a molecule of oxygen, Hb undergoes a conformational change that makes the binding of subsequent O2 molecules easier.
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3 years ago

The reaction A( g ) ⇌ 2 B( g ) A(g) ⇌ 2 B(g) has an equilibrium constant of K = 0.010 K = 0.010. What is the equilibrium constan

Over [174]

Answer:

K = 10

Explanation:

Using Hess's law, it is possible to obtain the equilibrium constant, K, of a reaction using K of similar reactions. For example:

B ⇄ A K = 1/X

2A ⇄ 2B K = X².

Thus, if A(g) ⇄ 2B(g) K = 0.010

2B(g) ⇄ A(g) K = 1 / 0.010; K = 100

B(g) ⇄ A(g) K = √100 = 10

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3 years ago

2. (10 Points) Several fast-acting medications are sold as powders, sometimes

Mila [183]

The fact that you want to yes is funny

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3 years ago

A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

bagirrra123 [75]

Answer : The percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.8870 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.8870-x)g}{74.5g/mole}=\frac{(0.8870-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.8870-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.142g}{143.32g/mole}=0.0149moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

0.0149 mole = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.818g$

The mass of NaCl = x = 0.818 g

The mass of KCl = (0.8870 - x) = 0.8870 - 0.818 = 0.069 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.818g}{0.8870g}\times 100=92.22\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.069g}{0.8870g}\times 100=7.78\%$

Therefore, the percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

3 0

3 years ago