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2 years ago

True or false: Pollen contains the male gamete of the plant

1 answer:

6 0

False pollen is transported to nearby plants by raindrops. ... embryo sac of angiosperms consists of eight diploid cells and one haploid egg cell

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In a titration experiment, 31.4 mL of 1.120 M HCOOH is neutralized by 16.3 mL of Ba(OH)2. What is the concentration of the Ba(OH

11Alexandr11 [23.1K]

Answer : The concentration of the $Ba(OH)_2$ solution is, 2.16 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCOOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$ .

We are given:

$n_1=1\\M_1=1.120M\\V_1=31.4mL\\n_2=2\\M_2=?\\V_2=16.3mL$

Putting values in above equation, we get:

$1\times 1.120M\times 31.4mL=2\times M_2\times 16.3mL\\\\M_2=2.16M$

Thus, the concentration of the $Ba(OH)_2$ solution is, 2.16 M

7 0

3 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

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2 years ago

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Would the amount of solid magnesium increase decrease or stay the same if you added a strong acid to it?

Vlada [557]

Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

so the amount of solid magnesium decrease by addition of strong acid to it.

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2 years ago

Based on the formulas, select the compounds that are covalent.

Sedbober [7]

S2F10, FE3N2, S7O2, NaCO3, Al2O3, PC13, and CO2. Hope this helps

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2 years ago

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How might society have been most likely impacted if Nikola Tesla, the inventor of AC technology, had not become a scientist?

Nonamiya [84]

Cars might not be as insulated and as safe because of the need to get cooler air into the vehicle, and im not fully sure but refrigerators would still run on ice to cool food. Houses wouldnt be as insulated from heat so when winter came around you would be colder than normal

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2 years ago

Read 2 more answers