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dem82 [27]
3 years ago
5

Please help!

Chemistry
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

You might be interested in
3. If 45.0 mL of 0.560 M KOH are needed to neutralize 125 mL of HCl of unknown concentration, what is
vlada-n [284]

Answer:

Molarity = 0.202 M

7.36 g/L HCl

Explanation:

KOH                        +                  HCl ----> KCl  +  H2O

1 mol                                           1 mol

0.560mol/L *0.045L                 x mol/L*0.125L

0.560mol/L *0.045L    =   x mol/L*0.125L

x = 0.560mol/L *0.045L  /0.125L= 0.2016 mol/L≈ 0.202 mol/L=0.202M HCl

M(HCl) = 1.0+35.5 = 36.5 g/mol

0.2016 mol/L*36.5 g/mol ≈ 7.36 g/L HCl

6 0
4 years ago
What is the identity of a sample that has a mass of 25.0 g and a volume of 2.38 cm3?
Snezhnost [94]
One of the defining features of an element is it's density. Calculate that and verify which substance this would be. 

p=m/v 
p=25g/2.38 cm^3 
p=10.50 g/cm^3 

Density of Silver is 10.5 g/cm^3 

Hope I helped :) 
7 0
4 years ago
Read 2 more answers
How many grams of FeCl3 are needed to make 1.5 L of a solution with a molarity of 0.450 M?
taurus [48]
Moles=concentration x volume. 1.5/1000 multiply 0.450 = 6.75x10^-4
To find the mass of fecl3, we have the formula, mass=moles x molar mass. 6.75x10^-4 x (55.8+35.5 x 3). = 0.1096g

Not sure tho if i did it right
6 0
3 years ago
Which of these solutes raises the boiling point of water the most?
Murljashka [212]

Answer:

D. ionic sodium phosphate (Na3PO4)

Explanation:

8 0
3 years ago
An igneous rock originally has 3 grams of uranium 238 in it. When dated the rock only contains 1.8 grams. What are the parent an
castortr0y [4]

Answer :

The parent and daughter concentrations (in percentages) is, 60 % and 40 % respectively.

The age of rock is 3.32\times 10^9\text{ years}

Explanation :

First we have to calculate the parent and daughter concentrations (in percentages).

\text{Parent concentrations}=\frac{1.8g}{3g}\times 100=60\%

and,

\text{Daughter concentrations}=\frac{(3-1.8)g}{3g}\times 100=40\%

As we know that, the half-life of uranium-238 = 4.5\times 10^9 years

Now we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{4.5\times 10^9\text{ years}}

k=1.54\times 10^{-10}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.54\times 10^{-10}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant  = 3 g

a - x = amount left after decay process = 1.8 g

Now put all the given values in above equation, we get

t=\frac{2.303}{1.54\times 10^{-10}}\log\frac{3}{1.8}

t=3.32\times 10^9\text{ years}

Therefore, the age of rock is 3.32\times 10^9\text{ years}

5 0
3 years ago
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