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2 years ago

The combining power of an element is called ?

Chemistry

1 answer:

Gre4nikov [31]2 years ago

8 0

Answer:

Explanation:

Valency is the combining power of an element

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In a neutralization reaction, an aqueous solution of an Arrhenius acid reacts with an aqueous solution of an Arrhenius base to p

solmaris [256]

Answer: c. Salt and Water

Explanation:

For example;

When an Arrhenius acid such as; Tetraoxosulphate (VI) acid (H2SO4) reacts with an Arrhenius base such as Potassium hydroxide (KOH), the products formed in this neutralization reaction is a salt known as ''Potassium Sulphate'' (K2SO4) and ''Water'' (H2O).

H2SO4 + KOH -------------> K2SO4 + H2O

5 0

3 years ago

Which of the following claims are based on scientific evidence? Red sky at night sailors' delight, red sky in morning sailors ta

vodomira [7]

I would believe that the answer to this question would be the "Touching toads gives you warts

3 0

3 years ago

Remember to balance the equations first

Semenov [28]

mol CO₂ = 9.6

mol N₂ = 4.8

mol O₂ = 0.8

mol H₂O = 8

<h3>Further explanation</h3>

Reaction

4C₃H₅O₉N₃ → 12CO₂ + 6N₂ + O₂ + 10H₂O

mol CO₂

$\tt \dfrac{12}{4}\times 3.2=9.6$

mol N₂

$\tt \dfrac{6}{4}\times 3.2=4.8$

mol O₂

$\tt \dfrac{1}{4}\times 3.2=0.8$

mol H₂O

$\tt \dfrac{10}{4}\times 3.2=8$

8 0

3 years ago

Write the symbol for the missing reactant.

Tcecarenko [31]

There is a shortcut trick while doing such fill in the blanks of nuclear reactions of hydrogen and helium

Let a,b,care elements of set N

$\\ \rm\Rrightarrow {}^{a}_1 H+{}^b_1H={}^c_2He\dots$

Here c must be a+b+1

Now

for our question

Hence b=4-3+1=1+1=2

The missing place should b e deuterium of heavy water

In nuclear reactions energy is released so it's mentioned on product side not reactant side

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2 years ago

1.2 kg of aluminum at 20oC is added to 1.5 kg of water at 80oC. After the system reaches thermal equilibrium, what is its final

Lubov Fominskaja [6]

Answer:

The final temperature is 71.19 °C

Explanation:

Step 1: Data given

Mass of aluminium = 1.2 kg = 1200 grams

Temperature of aluminium = 20.0 °C

Specific heat of aluminium = 0.900 J/g°C

Mass of water = 1.5 kg = 1500 grams

Temperature of water = 80.0 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the final temperature

heat gained = heat lost

Q(aluminium) = - Q(water)

Q = m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(water)

⇒ with mass of aluminium = 1200 grams

⇒ with specific heat of aluminium = 0.900 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 20.0 °C

⇒ with mass of water = 1500 grams

⇒ with specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 80.0°C

1200 * 0.900 * (T2-20.0°C) = -1500 * 4.184 * (T2 - 80.0°C)

1080 * (T2 - 20.0°C) = -6276 * (T2 - 80.0°C)

1080 T2 - 21600 = -6276T2 + 502080

7356T2 = 523680

T2 = 71.19 °C

The final temperature is 71.19 °C

8 0

3 years ago