All categories

3 years ago

Which of these is an organism? a virus b heart c bacteria d lung

Chemistry

2 answers:

worty [1.4K]3 years ago

7 0

Bacteria, because it is a living system :)

Nana76 [90]3 years ago

4 0

Hi , bacteria is the answer.

You might be interested in

How were atomic models developed when no one had seen an atom?

AnnyKZ [126]

Atomic models were developed through indirect observation even though no one had seen an atom. There were many experiments of which I know nothing about, but in the end the scientists managed to come up with a formula and various models to describe atoms.

4 0

3 years ago

Theory that the lithosphere is divided into tectonic plates that slowly move on

vitfil [10]

Pute because I’m smart and I know the right answer

5 0

3 years ago

Problem Page A chemist measures the amount of bromine liquid produced during an experiment. He finds that of bromine liquid is p

masha68 [24]

The question is incomplete, here is the complete question:

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.

Answer: The amount of liquid bromine produced is 4.79 moles.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of liquid bromine = 766. g

Molar mass of liquid bromine, $(Br_2)$ = 159.8 g/mol

Putting values in above equation, we get:

$\text{Moles of liquid bromine}=\frac{766.g}{159.8g/mol}=4.79mol$

Hence, the amount of liquid bromine produced is 4.79 moles.

6 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

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

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

2 years ago

Choose the letter of the choice that best completes the statement or answers the question. Elements in the same group have the s

pav-90 [236]

Elements in the same group have D. Same number of valence electrons.

6 0

3 years ago

Read 2 more answers