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Shtirlitz [24]
3 years ago
15

How many centimeters are there in 35,000 micrometers? Please and thank you

Chemistry
2 answers:
monitta3 years ago
5 0

Answer:

The answer is 3.5 centimeters.

Harman [31]3 years ago
3 0

Answer:

3.5

Explanation:

I looked it up

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Calculate the mass of oxygen in 30 g of CH NH COOH?​
Naya [18.7K]

Molar mass of CH2NH2COOH - 75

Given mass of CH2NH2COOH - 30

Moles of CH2NH2COOH = Given mass/ Molar mass

moles of CH2NH2COOH = 30/75 = 0.4 mol

One mole of CH2NH2COOH contains 32 gram of oxygen

0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen

Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!

7 0
3 years ago
Which or these equations is balanced H2SO4 + 2Al > Al2(SO4)3 + H2 or 2KCl + Pb(NO3)2 > 2KNO3 + PbCl2
Dmitry_Shevchenko [17]

Balancing of chemical equation is essential because of the law of conservation of mass, which states that the mass of a system can not be created or removed.

The second equation is balanced

2KCl_(_a_q_) +Pb(NO_3)_2_(_a_q_) ==> 2KNO_3_(-a_q_) + PbCl_2(_a_q_)

This is because the number of elements of each atom in the product side equal the number of elements of each atom on the reactant side.

The first equation is not balanced

H_2SO_4 _(_a_q) + 2Al_(_s_) ==> Al(SO_4)_3_(_a_q_) + H_2_(_g_)

This is because there is 1 molecule of SO_4^{-2} on reactant side as compared to 3 molecules of SO_4^{-2}

To balance the equation we add a coefficient of 3 on sulphuric acid (H_2SO_4) and a coefficient of 3 on hydrogen (H_2)

3H_2SO_4_(_a_q_) + 2Al_(_s_) ==> Al(SO_4)_3_(_a_q_) + 3H_2_(_g_)

7 0
4 years ago
From what you've learned so far, how is molecular structure related to smell?
Sidana [21]

Answer:

Changing the shape of the molecules that create fragrances in a flower or fruit may influence our perception of their smell. The reaction pattern produced, olfactory code, is sent as a signal to the brain, which which is how you smell things.

Explanation:

7 0
3 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0
3 years ago
Nitrogen is made up of two nitrogen atoms. Therefore, nitrogen is a
In-s [12.5K]
The chemical element of atomic number 7, a colorless, odorless unreactive gas that forms about 78 percent of the earth's atmosphere.
5 0
3 years ago
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