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3 years ago

Light energy, 6H2O, and 6CO2 are at the start of photosynthesis. True or False

Chemistry

2 answers:

IrinaVladis [17]3 years ago

4 0

The answer to this question is true

Kay [80]3 years ago

3 0

The best answer between the two choices would be the first option TRUE because light energy, 6H2O, & 6CO2 ARE at the start of photosynthesis.

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On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to t

Ne4ueva [31]

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is .........

Of F or Se, the larger atom is ..........

Of Sn or I, the larger atom is .........

Of Cr or W, the larger atom is ........

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.

Of Ge or Po, the larger atom is Po.

Of F or Se, the larger atom is Se.

Of Sn or I, the larger atom is I.

Of Cr or W, the larger atom is W.

7 0

3 years ago

Some commercially available algaecides for swimming pools claim to contain 7% copper. Could the method used in this experiment t

AleksandrR [38]

Answer:

Explanation:

7% copper implies 7 w/v%(weight/volume %) of copper. This implies a 100 mL algaecide arrangement contains 7 grams of copper.
Henceforth we have a thought of the measure of copper that ought to be available in a given example of algaecide.
Indeed, even a 1 mL algaecide test is sufficient to discover the percentage of copper in it.
so a 25mL sample of algaecide must have around 1.75g of copper.

4 0

3 years ago

The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,

Katyanochek1 [597]

Answer:

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

$n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

$n_g = 0.00215 \ mols$

Given that the number of moles of CHCl3 evaporated is less than the number of moles of CHCl3 initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of air so the pressure at equilibrium is 100 mmHg

4 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago

Consider the reaction:

kramer

Answer:

SOMEONE ANSWER THIS PLEASE

Explanation:

PLEASE

3 0

2 years ago