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natta225 [31]
3 years ago
12

Which is a TRUE statement about salt water?

Chemistry
2 answers:
makkiz [27]3 years ago
6 0
I think the best answer from the given choices is option D. <span>Salt water is made up of more than one pure substance. It consists of salt and water. When each component exist alone in a container they are considered as pure substance however when mixed they are now a solution.</span>
Phantasy [73]3 years ago
3 0

Answer:

the answer is D

Explanation:

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What is combination reaction and also define it's properties?​
tigry1 [53]

Answer:

This is when two or more compounds or elements react to form a single product.

In these reactions new substances are produced or synthesized.

in simple terms the elements or compounds are converted to something new.

5 0
3 years ago
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_____ NaPO4+_____KOH
Artist 52 [7]
NaPO4 + KOH -> KPO4 + NaOH
already balance
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3 years ago
A sample of an ideal gas at 1.00 atm and a volume of 1.32 L was placed in a weighted balloon and dropped into the ocean. As the
Amiraneli [1.4K]

Answer:

When the pressure increases to 90.0 atm , the volume of the sample is 0.01467L

Explanation:

To answer the question, we note that

P₁ = 1.00 atm

V₁ = 1.32 L

P₂ = 90 atm.

According to Boyle's law, at constant temperature, the volume of gas is inversely proportional to its pressure

That is P₁V₁ = P₂V₂

Solving the above equation for V₂ we have

V_2 = \frac{P_1P_2}{P_2}  that is V₂ = \frac{1atm*1.32L}{90atm} = \frac{11}{750}L or 0.01467L

5 0
3 years ago
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An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The
GuDViN [60]

Answer:Mass of Potassium chloride =1.762g

Explanation:

Mass of empty beaker = 23.100 g

Mass of beaker with Potassium chloride = 24.862g

Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride  - Mass of empty beaker = 24.862-23.100 = 1.762g

8 0
3 years ago
Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
3 years ago
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