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3 years ago

If I have 4 moles of gas at a pressure of 7.6 atm and a volume of 19 liters, what is the temperature?

Chemistry

1 answer:

lesantik [10]3 years ago

3 0

Answer:

P = 114.3722951 torr.

Explanation:

Assuming an ideal gas, PV = nRT. Since you're dealing with torr, we should use 62.4 as our R constant.

Thus:

488P = 3*62.4*(273.15+25)

Solving for P:

P = 114.3722951 torr.

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What is the most reactive metal

ohaa [14]

Some people think its francium and some people believe it is either sodium or potassium.

Some people believe that the most reactive element is francium because the larger the atom, the easier it is to get rid of the valence electrons and the easier it is to react with other atoms. so now if you look at the periodic table, francium has the largest radius and so that's why some people think its francium. but in my opinion, Francium AND Cesium.

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2 years ago

An unknown substance has a boiling point of 174°C. Which statement is true about this unknown substance?

sertanlavr [38]

Answer:

Explanation:

8 0

2 years ago

Read 2 more answers

Assume that the NO, concentration in a house with a gas stove is 150 pg/m°. Calculate the equivalent concentration in ppm at STP

jok3333 [9.3K]

Explanation:

It is known that for $NO_{2}$ , ppm present in 1 $mg/m^{3}$ are as follows.

1 $\frac{mg}{m^{3}}$ = 0.494 ppm

So, 150 $pg/m^{3}$ = $\frac{150}{1000} mg/m^{3}$

= 0.15 $mg/m^{3}$

Therefore, calculate the equivalent concentration in ppm as follows.

$0.15 \times 0.494 ppm$

= 0.074 ppm

Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.

4 0

2 years ago

Explain why hydrogen can only have 2 valence electrons around it when it bonds to other items. What is the maximum number of bon

lys-0071 [83]

Answer:

In first shell only 2 electrons are present in hydrogen only one electron is present in valence shell. To complete its duplet hydrogen can share lose or gain only one electron to form chemical bond.

Hydrogen can form only one bond.

Explanation:

7 0

2 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

2 years ago