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ryzh [129]
3 years ago
10

What type of bonding is found in the molecule HBr?

Chemistry
1 answer:
Liula [17]3 years ago
4 0
D)polar covalent because ionic and metallics include metals and nonpolar covalent wouldn’t work because it’s lewis for structure and en doesn’t match its requirements.
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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
Which represents the balanced nuclear equation for the beta plus decay of C-11?
slavikrds [6]
Nuclear reaction: ¹¹C → ¹¹B + e⁺(positron) + ve(electron neutrino).<span><span><span><span>
</span></span></span></span>Beta decay is radioactive decay<span> in which a beta ray and a neutrino are emitted from an atomic nucleus.
There are two types of beta decay: beta minus and beta plus. In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino and in beta plus decay, a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.</span>
8 0
3 years ago
Read 2 more answers
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Nuetrik [128]

Explanation:

What's Dis Suppose To Mean ?

7 0
3 years ago
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at
lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

6 0
2 years ago
2C 2 H 6 +7O 2 ***&gt;4CO 2 +6H 2 O if 7.0 g of C 2 H 6 react with 18 g of O 2 , how many grams of water will be produced
Alex787 [66]

Answer:

grams H₂O produced = 8.7 grams

Explanation:

Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)

               7g           18g                             ?g

Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water

Moles Reactants

moles C₂H₆ = 7g/30g/mol = 0.233mol

moles O₂ = 18g/32g/mol = 0.563mol

Limiting Reactant => (Test for Limiting Reactant)  Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.

moles C₂H₆/2 = 0.233/2 = 0.12

moles O₂/7 = 0.08

<u><em>Limiting Reactant is O₂</em></u>

Moles and Grams of H₂O:

Use Limiting Reactant moles (not division value) to calculate moles of H₂O.

moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield

grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O

3 0
3 years ago
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