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3 years ago

PLEASE IN NEED HELP ASAP

1 answer:

3 0

Because nothing appeared he concluded that that living things could NOT just appear or come from nonliving things like water and air

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¿Cuál será la masa molar del agua?, si el hidrógeno tiene una masa atómica H=1.0 uma y el Oxigeno de 16 uma. * 1 punto

nataly862011 [7]

Answer:

Explanation:

Sabemos que la fórmula química del agua es H2O.

Usando la información que; H = 1, O = 16, podemos calcular la masa molar del agua de la siguiente manera

[2 (1) + 16] = 18

Por tanto, la masa molar del agua es 18

3 0

3 years ago

PLEASE HELP IMMEDIATELY I NEED THE ANSWER NOT A HINT THANK YOU

Tcecarenko [31]

Cytoplasm: gel like environment which allows organelles to move about the cell

Golgi bodies: packages and ships materials out of the cell

Cell membrane: controls what goes in and out of the cell

Chloroplast: makes food for plant cells using sunlight

Lysosome: breaks down waste, food, and worn out cell parts

Mitochondria: breaks down food to release energy for the cell

Nucleus: contains the cell's DNA and is the control center of the cell

Endoplasmic reticulum: transports materials within cell; process lipids

Vacuole: stores water, waste and food

Ribosome: make proteins

5 0

3 years ago

What happens to a light wave that travels through glass and strikes a

Mademuasel [1]

Answer:D

it reflects back into the air moving the opsite direction.

i hope this helps :)

8 0

3 years ago

Which of these shows the correct hierarchical sequence?

gayaneshka [121]

Answer:

cells -> tissues -> organs -> organ system

4 0

3 years ago

Read 2 more answers

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

$\frac{44}{56}$

= 0.785 g of CO2

So,

14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g of CaO produce =

$\frac{100}{56}$

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

$\frac{Actual\ yield}{Theoritical\ yield}\times 100$

$\frac{19.4}{25.71}\times 100$

= 75.44 %

6 0

3 years ago