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2 years ago

In an atom the number of protons must be equal of

Chemistry

2 answers:

Svetlanka [38]2 years ago

8 0

Answer:

electrons

Explanation:

Tju [1.3M]2 years ago

4 0

Answer:

Electrons

Explanation:

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What's the percentage of ZnF2

Elza [17]

I don’t understand the question

6 0

3 years ago

The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr - Cp*mu/k where Cp is the heat

algol [13]

Answer:

The Prandtl number for this example is 14,553.

Explanation:

The Prandlt number is defined as:

$Pr=\frac{C_{p}*\mu}{k}$

To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.

$\mu=1896 \frac{lbm}{ft*h}*\frac{1000 g}{2.205 lbm}*\frac{3.281 ft}{1 m}*\frac{1h}{3600s} =7938 \frac{g}{m*s}$

Now that we have coherent units, we can calculate Pr

$Pr=\frac{C_{p}*\mu}{k}=0.66*7938/0.36=14553$

8 0

3 years ago

A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A

Gemiola [76]

Answer:

(0,653±0,002) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =

0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,653±0,002) M of HNO₃

I hope it helps!

5 0

3 years ago

Why does fire burn wood

Maurinko [17]

Answer:

Because of how dry it is

Explanation:

5 0

3 years ago

Copper(II) fluoride contains 37.42% F by mass. Calculate the mass of fluorine (in g) in 55.5 g of copper(II) fluoride.

Mrac [35]

Answer:

There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride

Explanation:

x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen

37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.

So, 100 g of copper (II) fluoride contains 37.42 g of F

55.5 g of copper (II) fluoride contains $\frac{37.42\times 55.5}{100}$ g of F or 20.8 g of F

Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.

5 0

3 years ago