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klemol [59]
4 years ago
11

Observing a phenomenon in the lab includes which of the following?

Chemistry
1 answer:
dedylja [7]4 years ago
8 0

Answer:

D

Explanation:

Identifying unique features of evidence

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Help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
nata0808 [166]

Answer:

1.

  • A) protons and neutrons
  • B) electrons

2.

  • By changing the number of proton.

Explanation:

Protons and neutrons are located at the center of an atom which is nucleus .

Electrons will orbit around the nucleus .

3 0
3 years ago
Read 2 more answers
How many grams of a 19.6% sugar solution contain 72.5 g of sugar?
anzhelika [568]

Answer:

6.2g of sugar solution contains 72.5g

4 0
3 years ago
Magnesium reacts with hydrochloric acid (HCl) as follows. mc025-1.jpg How many milliliters of hydrogen gas are produced by the r
Juliette [100K]

Answer: 448 mL of hydrogen gas.

Solution:

Mg(s)+HCl(aq)\rightarrow MgCl_2(aq)_+H_2(g)

Now,according to reaction

1 mole of magnesium produces one mole of hydrogen gas.

Then 0.020 moles of magnesium will produce 0.020 moles of hydrogen gas.

Moles of H_ gas : 0.020 mol

At STP, 1 mol of gas occupies 22.4 L

So volume of 0.020 mol of H_2produced at STP :

= 22.4/times 0.020 = 0.448 L = 448 mL    (1L=1000mL)

8 0
4 years ago
Read 2 more answers
When Earth’s axis points away from the sun, the _________ Hemisphere has the longer days of summer.
Llana [10]

Answer: northern hemisphere

Explanation: I looked it up. Plus I took a test with this question and when the teacher went over the answer i got it right.

3 0
3 years ago
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
lions [1.4K]

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

4 0
3 years ago
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