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3 years ago

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In the following reaction, what coefficient will be written before the potassium nitrate (KNO3 ) to balance the chemical equatio

n? (1 point) _____ K2CrO4 + _____ Pb(NO3)2 _____ KNO3 + _____ PbCrO4

1 answer:

puteri [66]3 years ago

4 0

Hey there:

<span>Balanced chemical equation:
</span>
<span>1 K2CrO4 + 1 Pb(NO3)2 = 2 KNO3 + 1 PbCrO<span>4
</span></span>
<span>Coefficients : 1 , 1 , 2 , 1
</span>
hope this helps!

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A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )

Carbon

$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

$n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }$

$n_{C} = 0.504\,moles$

Oxygen

$n_{O} = \frac{r\cdot m_{O}}{M_{O}}$ (3)

$n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }$

$n_{O} = 1.008\,moles$

Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

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3 years ago

is 0.0410 M−1 s −1 . We start with 0.105 mol C2F4 in a 4.00-liter container, with no C4F8 initially present. What will be the co

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Answer:

After three hours, concentration of C₂F₄ is 0.00208M

Explanation:

The rate constant of the reaction:

2 C2F4 → C4F8 is 0.0410M⁻¹s⁻¹

As the units are M⁻¹s⁻¹, this reaction is of second order. The integrated law of a second-order reaction is:

$\frac{1}{[A]} =\frac{1}{[A]_0} +Kt$

<em>Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time of the reaction (In seconds). </em>

3.00 hours are in seconds:

3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds

Initial concentration of C2F4 is:

0.105mol / 4.00L = 0.02625M

Replacing in the integrated law:

$\frac{1}{[A]_0}= \frac{1}{0.02625} +0.0410M^{-1}s^{-1}*10800s\\\frac{1}{[A]_0}=480.9M^{-1}$

[A] = 0.00208M

<h3>After three hours, concentration of C₂F₄ is 0.00208M</h3>

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3 years ago

For which type of titration will the ph be basic at the equivalence point?

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Answer:

weak acid

Explanation:

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4. The blood vessels, heart and the blood make up

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Answer:

D, the cardiovascular system

Explanation:

The Cardiovascular System is the system at which blood circulates through our body

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3 years ago

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Chlorine consists of two isotopes with masses of 35 and 37. If the average atomic mass of a sample of chlorine atoms is 35.5, wh

Elza [17]

Answer:

The ratio 35Cl/37Cl = 3/1

Explanation:

<u>Step 1:</u> Data given

Chlorine has 2 isotopes:

- mass = 35 g/mol

- mass = 37 g/mol

Average molar mass of chlorine = 35.5 grams

<u>Step 2: </u>Calculate the % of isotopes

35x + 37y = 35.5

x+y = 1 or x = 1-y

35(1-y) + 37y = 35.5

35-35y +37y = 35.5

0.5 = 2y

y = 0.25 = 37Cl

x = 1 - 0.25 = 0.75 = 35Cl

<u>Step 3: </u>

The ratio 35Cl/37Cl = 0.75/0.25 = 3/1

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4 years ago