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Vesnalui [34]
2 years ago
13

What are body parts that have lost their original function through evolution?

Chemistry
1 answer:
katrin2010 [14]2 years ago
6 0

Answer:

Vestigial structures are body parts that have lost their use through evolution.

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How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac
Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

4 0
1 year ago
When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal
Mnenie [13.5K]

Answer:

A is denser than B as it's volume for the same mass is smaller.

Explanation:

Hello.

In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:

V_A=50.0mL-38.0mL=12.0mL\\\\V_B=60.0mL-38.0mL=22.0mL

Next, we compute the densities as shown below:

\rho_A=\frac{m_A}{V_A}=\frac{5.00g}{12.0mL}=0.42g/mL\\  \\\rho_B=\frac{m_B}{V_B}=\frac{5.00g}{22.0mL}=0.23g/mL

In such a way, A is denser is B as it's volume for the same mass is smaller.

Best regards.

4 0
3 years ago
What is the affect of increasing the water's mass?how does it reflect it's temperature?
Nikolay [14]
I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ? 
Well the formula is : Q=m\cdot c\cdot \Delta t (where Q is heat, m is mass, c is specific heat and \Delta t is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless,  if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature
8 0
3 years ago
There are two open cans of soda on the table. One can was just taken from the refrigerator and the other was taken from the cupb
nika2105 [10]
I think the correct answer from the choices listed above is option C. The can <span>from the cupboard will lose carbon dioxide more quickly because it is warmer and gases are less soluble in warmer temperatures. </span> Solubility of gases is a strong function of temperature and as well as pressure.
5 0
3 years ago
Read 2 more answers
The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,
Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by t_{\frac{1}{2}

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. \frac{2400}{2}=1200, after second  half life, the activity would be reduced to half of 1200 i.e. \frac{1200}{2}=600,  and after third half life, the activity would be reduced to half of 600 i.e. \frac{600}{2}=300,

Thus the activity 9 minutes later is 300 bq.

7 0
3 years ago
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