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2 years ago

Please help me I really need it :(

Chemistry

2 answers:

masya89 [10]2 years ago

8 0

1)If the temperature of the water changes then the amount of sugar that can dissolve changes because hot water makes more amount of sugar dissolve.

2)If the amount of gas in a car changes then the distance a car can travel doesn’t change because the amount of gas doesn’t affect the distance a car can travel.

this is in 7th grade form so it might not be good if you’re in a higher grade

skad [1K]2 years ago

7 0

Answer: i Honestly don’t know

Explanation:

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Theoretically, how many moles of carbonic acid will be produced by 3.00 g sample of NaHCO3?

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Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27%

3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3

0.8211 grams Na + 1.266 grams Cl = 2.087 grams

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the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh

Harman [31]

Answer:

$\boxed{\text{10 L}}$

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

$\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}$

Calculation:

$\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}$

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Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are: Cl⁻, Na⁺ (Don't react, spectator ions), and OH⁻.

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