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3 years ago

What is a common ancestor?

Chemistry

2 answers:

yawa3891 [41]3 years ago

7 0

A Common ancestor is a organism that has shared predecessor of more then one different scion of a group of organisms.

Agata [3.3K]3 years ago

6 0

In evolutionary biology, a common ancestor is allegedly a link between two varying branches or types of creatures. (Another word for these is 'missing links', since up to this point, not a single one of these common ancestors has been found in the fossil record)
Hope that helped =)

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A chemist has a 12.5 liter sample of neon gas at 1.00 atm and 30 c. If the chemist compresses the sample to 10.5 liters and hold

ycow [4]

Answer:

254.5 K

Explanation:

Data Given

initial volume V1 of neon gas = 12.5 L

final Volume V2 of neon gas = 10.5 L

initial Temperature T1 of neon gas = 30 °C

convert Temperature to Kelvin

T1 = °C +273

T1 = 30°C + 273 = 303 K

final Temperature T2 of neon gas = ?

Solution:

This problem will be solved by using Charles' law equation at constant pressure.

The formula used

V1 / T1 = V2 / T2

As we have to find out Temperature, so rearrange the above equation

T2 = V2 x T1 / V1

Put value from the data given

T2 = 10.5 L x 303 K / 12.5 L

T2 = 254.5K

So the final Temperature of neon gas = 254.5 K

8 0

3 years ago

Please help me find the final answer

dlinn [17]

Answer:

5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃

6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693. $\overline{3168}$ grams of N₂

Explanation:

5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃) is presented as follows;

(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O

One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂

The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;

$\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}$

$\therefore {x \ g} = \dfrac{5 \ g}{44.01 \ g} \times 105.99 \ g \approx 12.04 \ g$

The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g

6. The chemical equation for the reaction is presented as follows;

N₂ + 3H₂ → 2NH₃

Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)

The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;

$\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}$

$\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g$

The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693. $\overline{3168}$ grams

6 0

3 years ago

1.40 m3 is how many mL

Zarrin [17]

$\LARGE{ \boxed{ \rm{ \pink{Solution:}}}}$

We know, 1 m³ of space can hold 1000 l of the substance.

⇛ 1 m³ = 1000 l----(1)

And, 1 l is 1000 times more than 1 ml

⇛ 1 l = 1000 ml------(2)

So, From (1) and (2),

⇛ 1 m³ = 1000 × 1000 ml

⇛ 1m³ = 1000000 ml

We had to find,

⇛ 1.40 m³ = 1.40 × 1000000 ml

⇛ 1.40 m³ = 140/100 × 1000000 ml

⇛ 1.40 m³ = 1400000 ml

⇛ 1.40 m³ = 14,00,000 ml / 14 × 10⁵ ml / 1.4 × 10⁶ ml

☃️ So, 1.40 m³ = 14 × 10⁵ ml / 1.4 × 10⁶ ml.

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5 0

3 years ago

Read 2 more answers

352 mL of chlorine exerts a pressure of 680 mmHg. What will the volume

Drupady [299]

Answer:

The answer would be 0.198 liters.

Explanation:

Message me for any questions.

7 0

3 years ago

Help plz:)))I’ll mark u Brainliest

emmasim [6.3K]

Answer:

$\Delta _{fus}H=205J/g=13.03kJ/mol$

Explanation:

Hello there!

In this case, since the heat of fusion is a property that allows us to calculate the heat involved during the change from solid to liquid (fusion) and is calculated as shown below:

$Q=m*\Delta _{fus}H$

In such a way, given the heat involved during this process and the mass of copper, we calculate the heat of fusion as shown below:

$\Delta _{fus}H=\frac{Q}{m}=\frac{41000J}{200.g}\\\\\Delta _{fus}H=205J/g$

Or in kJ/mol:

$\Delta _{fus}H=205\frac{J}{gCu}*\frac{63.546 gCu}{1molCu}\\\\ \Delta _{fus}H=13026.93J/mol=13.03kJ/mol$

Regards!

8 0

3 years ago