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irga5000 [103]
3 years ago
11

At the start of the _________, about 10,000 years ago, humans began planting crops and domesticating animals.

Chemistry
1 answer:
Alex777 [14]3 years ago
5 0
I think the Neolithic Revolution is the correct answer
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Cuantos gramos de nitrato de potasio (KNO3) son necesarias para preparar 800ml de solucion al 12 % m/v
maxonik [38]

Answer:

96 g

Explanation:

Tomando en cuenta que estamos hablando de una solución de Nitrato de potasio cuya concentración es 12% m/V, debemos entonces usar la expresión de %m/V:

%m/V = msto / Vsol * 100   (1)

De esta expresión, debemos despejar la masa de soluto (msto) que es precisamente el valor que estamos buscando. Al hacer el despeje tenemos:

msto = %m/V * Vsol / 100    (2)

Ahora solo nos queda sustituir los valores dados y resolver:

msto = 12 * 800 / 100

<h2>msto = 96 g de KNO₃</h2>

Espero te sirva.

3 0
3 years ago
What is physiography ? ​
Alexxandr [17]

Answer:

Physical geography is one of the two fields of geography. Physical geography is the branch of natural science which deals with the processes and patterns in the natural environment such as the atmosphere

8 0
3 years ago
You have 500,000 atoms of a radioactive substance. After 2 half-lives have past, how many atoms remain?
Aleonysh [2.5K]

Answer:

125000

Explanation:

Because it is halved and halved again.

3 0
2 years ago
Read 2 more answers
Which of the following leads to a higher rate of diffusion?
Bingel [31]

I thought the answer is d

5 0
3 years ago
Read 2 more answers
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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