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skad [1K]
3 years ago
10

Think about the typical weather conditions where you live. What is the name of the average yearly weather patterns in any one ar

ea?
Chemistry
1 answer:
Ludmilka [50]3 years ago
5 0

CLIMATE

give me brainliest please

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Why bond angle of H2O is maximum then OF2??
Nataly_w [17]
It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.
8 0
3 years ago
How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g

Therefore, the mass of PbSO_4 needed are, 1.515 grams.

6 0
3 years ago
How many photons are produced in a laser pulse of 0.364 J at 477 nm?
Luda [366]

Answer:

1.00 × 10¹⁸

Explanation:

1. Calculate the <em>energy of one photon</em>

The formula for the energy of a photon is

<em>E</em> = <em>hc</em>/λ  

<em>h</em> = 6.626 × 10⁻³⁴ J·s; <em>c</em> = 2.998 × 10⁸ m·s⁻¹

λ = 477 nm = 477 × 10⁻⁹ m                              Insert the values

<em>E</em> = (6.626 × 10⁻³⁴ × 2.998× 10⁸)/(477 × 10⁻⁹)

<em>E</em> = 4.165× 10⁻¹⁹ J

2. Calculate the <em>number of photons</em>

Divide the total energy by the energy of one photon.

No. of photons = 0.418 × 1/4.165 × 10⁻¹⁹      

No. of photons = 1.00 × 10¹⁸

7 0
4 years ago
Which is mostly likely to cause clouds to form in a previously clear sky?​
xxTIMURxx [149]

Answer:

Curious clouds.... ...

6 0
3 years ago
Read 2 more answers
What is similar about the outer shell of electrons in the alkali metals family
Nadusha1986 [10]

Answer:

1 electron

Explanation:

These metals have a single electron in the outer shell

4 0
3 years ago
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