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2 years ago

What is a oganisem mmmmmmmm

Chemistry

2 answers:

Gemiola [76]2 years ago

7 0

Living things that are capable of reacting to stimuli, reproduction, growth, and homeostasis.

leva [86]2 years ago

7 0

Answer:

an individual animal, plant, or single-celled life form.

Explanation:

Hope that helps:)

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DDT is currently banned worldwide. a. True b. False

Aleks [24]

B. False. It is banned worldwide for agricultural uses. The use of DDT is still authorized in small proportions in countries that need it, with support deployed for the alteration to safer and more effective alternatives.

5 0

3 years ago

What is the only force that can act on an object in free fall?

kkurt [141]

I’m going to say Gravity.

8 0

2 years ago

g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4

kykrilka [37]

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

$Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2$

Stoichiometrically, 1 mole of $Na_2C_2O_4$ reacts with 2 moles of $HCl$ for a complete neutralization reaction.

Recall that: mole = $\frac{mass}{molar mass}$

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole $Na_2C_2O_4$ requires 2 moles HCl, then 0.0041 mole will require:

0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

7 0

3 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

At standard pressure, during which physical change does the potential energy decrease?

skad [1K]

Gas - > liquid - > solid

6 0

3 years ago

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