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3 years ago

Which one is it (it’s multiple choice too)

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

5 0

Answer:

i think its b and c.

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HELP PLS 50 POINTSSSS

Alenkasestr [34]

Answer:

Carbon dioxide is a product (cellular respiration)

Carbon dioxide is a reactant(photosynthesis)

Carried out in animals(cellular respiration)

Carried out in plants(both)

Chemical reaction(cellular respiration)

Oxygen is a product(photosynthesis)

Oxygen is a reactant(cellular respiration)

Produces usable energy source(photosynthesis)

Your welcome!!! Plz mark brainliest.

4 0

3 years ago

Read 2 more answers

Which of the following is an

eimsori [14]

Answer:

2 and 4

Explanation:

The rest of the changes are chemical. 1 has a chemical reaction happen which makes light sticks glow. 3 is because browning the meat actually causes some new compounds to form and cause caramelization.

For 2, the oxygen is simply heating up and expanding which pops the balloon. 4 is just a phase change of water vapor to liquid water.

5 0

3 years ago

0.0075 mole of calcium trioxocarbonate(IV) is

KatRina [158]

Answer:

0.0075÷0.015=0.5 not so sure lol

4 0

4 years ago

Reactant(s) H 0, Product(s) H,0 3.4g 10g

zalisa [80]

Answer:

H,0

3.4g

Explanation:

H,0

3.4g

8 0

3 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago