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3 years ago

Please Help!!!!!!!!!!!!!!

Chemistry

2 answers:

defon3 years ago

8 0

Answer:

i can't see it

Explanation:

gladu [14]3 years ago

5 0

So wouldn’t you just type the scientific notation that it gave you into a calculator and then get the answer?

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typical seawarer contains 2.7g of salt (sodium chloride,Nacl)per 100ml (100*10-3L).what is the molarity of Nacl in the ocean?

faust18 [17]

Answer:

Molarity = 0.5 M

Explanation:

Given data:

Mass of NaCl = 2.7 g

Volume = 100 mL(100×10⁻³L)

Molarity of solution = ?

Solution:

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 2.7 g/ 58.5 g/mol

Number of moles = 0.05 mol

Molarity = number of moles / volume in litte

Molarity = 0.05 mol / 100×10⁻³L

Molarity = 0.5 M

4 0

3 years ago

Calcium carbonate, CaCO3(s), the principal compound in limestone, decomposes upon heating to CaO(s) and CO2(g). A sample of CaCO

lorasvet [3.4K]

Answer:

n = 0.01302 moles

Explanation:

Pressure = 1.3 atm

Temperature = 31 °C + 273 = 304 K ( Converting to kelvin temperature)

Number of moles = ?

Volume = 250 mL = 0.250 L

These quantities are related by the equation below;

PV = nRT

n = PV / RT

n = (1.3 * 0.250) / (0.0821 * 304)

n = 0.01302 moles

5 0

3 years ago

The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins

saul85 [17]

The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

So, we know:
t = 10 min
 $t_{1/2}$ = 2.5 min

We need:
n = ?
x = ?

We could first use the second equation to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{10 min}{2.5 min}$
⇒ $n=4$ 

Now we can use the first equation to calculate the remained fraction of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^4$
⇒ $x= \frac{1}{16}$

3 0

3 years ago

Fluidity is maximum in a. Solid b. Liquid c. Gas d. Liquid and gas

Phantasy [73]

The answer is C gas

8 0

3 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

3 years ago