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kolezko [41]
2 years ago
5

46.0 mL sample of gas at 28.0°C and 748 torr. What volume would this gas sample occupy at stp?

Chemistry
1 answer:
andreyandreev [35.5K]2 years ago
7 0

Answer:

2

Explanation:

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Sergio [31]

Answer:

C. The Protons

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3 years ago
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What is the name for SO2
Tanzania [10]
Sulphur Dioxide. Toxic. Don't eat it.
7 0
3 years ago
at a certain pressure the density of superficial carbon dioxide is 0.469. what is the mass of a 25.0 mL sample of supercritical
Murrr4er [49]

Answer:

\large \boxed{\text{11.7 g}}

Explanation:

\text{Mass} = \text{25.0 mL} \times \dfrac{\text{0.469 g}}{\text{1 mL}} = \textbf{11.7 g}\\\\\text{The mass of the carbon dioxide is $\large \boxed{\textbf{11.7 g}}$}

7 0
3 years ago
Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2
svp [43]

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S),         Spring Constant (N/m)

      W,                   24

      X,                    35

      Y,                    22

      Z,                    15

The equation for elastic potential energy, E_e, is E_e = 0.5 \times k \times x^2

The above equation can also be written as E_e =\dfrac{1}{2}  \times k \times x^2

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, E_e, of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, E_e, therefore the correct order is as follows;

X > W > Y > Z

9 0
3 years ago
What volume of Co2 (carbon (iv) oxide)
hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0
3 years ago
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