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3 years ago

7

The solubility for KCl is 60 g at 40 C. If a solution of KCl contains 79.5 g, is the solution saturated, unsaturated, or supersa

turated

1 answer:

Alekssandra [29.7K]3 years ago

7 0

Answer:

saturated

Explanation:

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Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr

Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

$0.0366 M=\frac{n}{5.00 L}$

$n=0.0366 M\times 5 mol=0.183 mol$

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

$n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}$

$0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}$

$106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}$

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0

3 years ago

Sodium (chemical symbol Na) is atomic number 11 and is in the first column of the periodic table so it has one valence electron.

lisabon 2012 [21]

Na is cation so it loses electron to be positive and become stable losing one valence shells one electron so it's oxidation number is +1 ie A is correct

7 0

4 years ago

Read 2 more answers

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0

2 years ago

In the sodium-potassium pump, sodium ion release causes two K ions to be released into the cell. __________________ True False

Solnce55 [7]

Answer:

false

Explanation:

As we know that in sodium-potassium pump .

sodium potassium move 3Na+ outside the cells

and moving 2k+ inside the cells

so that we can say that given statement is false

Answer FALSE

6 0

3 years ago

This data was collected after conducting an experiment about the amount, in liters, of water a specific plant needs per month. A

Effectus [21]

Answer:

1.1 liters

1.2 liters

1.5 liters

Explanation:

Precision in data refers to how close the experimental values of an experiment are to one another irrespective of the true or accepted value. In other words, a set of values are said to be PRECISE if they are close to one another.

In this case, data was collected after conducting an experiment about the amount, in liters, of water a specific plant needs per month. However, according to the set of experimental values provided, only 1.1 litres, 1.2litres and 1.5litres are close to one another and, hence, are said to be PRECISE even if they are not close to the accepted value of 6litres.

4 0

3 years ago

Read 2 more answers