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maria [59]
3 years ago
14

the absolute tempeature of a gas is increased four times while maintaining a constant volume. what happens to the pressure of th

e gas​
Chemistry
1 answer:
ikadub [295]3 years ago
6 0

Answer:

increases

Explanation:

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When 32 grams of aluminum react, the actual yield is 105.5 grams, what is the percent yield?
user100 [1]

Answer:

329.7%

Explanation:

Percent Yield = Actual Yield/ Theoretical Yield x 100%

Percent Yield = 105.5g/32 x 100% = 329.69 ≈ 329.7 %

5 0
3 years ago
Number 1 please fast answer
Natasha2012 [34]

Answer:

I will give you 5

Explanation:

color chage, formation of a precipitate, formation of a gas, odor change, temperature change. are all signs of a chemical reaction

3 0
2 years ago
Read 2 more answers
How much energy is required to raise the temperature of 10.6 grams of gaseous neon from
Alona [7]

Answer:

Approximately 1.95 \times 10^{2}\; \rm J.

Explanation:

Look up the specific heat of gaseous neon:

c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}.

Calculate the required temperature change:

\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K.

Let m denote the mass of a sample of specific heat C. Energy required to raise the temperature of this sample by \Delta T:

Q = c \cdot m \cdot \Delta T.

For the neon gas in this question:

  • c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}.
  • m = 10.6\; \rm g.
  • \Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K.

Calculate the energy associated with this temperature change:

\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}.

3 0
3 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
A gas has a volume of 56.3L when its temperature is 280 K. What will volume be at 220K?
Alex Ar [27]

Answer:

44.23 L

use pascal's formula

8 0
3 years ago
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