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3 years ago

How does the study of cosmology relate to astronomy

1 answer:

6 0

Cosmology is the study of the grand scale of the universe. You need astronomical observation because then there would be no proof to cosmological theory.

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Which feature do all adult echinoderms have?

Rama09 [41]

The common name to the family member of phylum Echinodermata of marine family is echinoderm. They are usually characterized by a five-fold symmetry, and possess an internal skeleton of calcite plates. They are found at every ocean depth.

The features of all adult echinoderm are:

- They have a five-fold symmetry.

- Body without segmentation.

- Spiny skin.

- Internal skeleton.

- found at every ocean depth.

4 0

3 years ago

Read 2 more answers

Jake collected samples of two substances while he was out walking. After taking the samples home, he ran tests and found that on

inysia [295]

The answer is: first substance is base and second is acid.

Bases, for example solution of sodium hydroxide (NaOH) feels slipery.

Sodium hydroxide dissociation in water: NaOH(aq) → Na⁺(aq) + OH⁻(aq).

Ionic compounds are good good electricity and heat conductors, because ionic compounds have mobile ions (cations and anions) that are able to transfer electrical charge.

In second reaction, magnesium is oxidized from oxidation number 0 to +2 and hydrogen is reduced from +1 to 0 (hydrogen gas).

7 0

4 years ago

Read 2 more answers

A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant

KiRa [710]

Answer:

$V_2=2.995L\\\\W=248.5J$

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

$P_1V_1=P_2V_2$

So we solve for $V_2$ by firstly computing the initial pressure:

$P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L} =11.98atm$

$V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L$

Finally, we can compute the work by using the following formula:

$W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J$

Best regards.

4 0

4 years ago

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago

Who discovered phosphate?

iris [78.8K]

Answer:

Hennig Brand

Explanation:

4 0

3 years ago