Atoms that have lost or gained one or more electrons are called _______

What is the concentration of each ion present in a saturated aqueous solution of the compound BaSO4. Ksp = 1.1 x 10^-10

wel

Answer:

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7 0

3 years ago

Which element would be the most reactive with K? Br, Na, Ar, or O? And how do you know?

kykrilka [37]

Ar I did this I think i got it right on edunuty

5 0

3 years ago

Im hvaing a hard time getting the right answer

Tanya [424]

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!

6 0

3 years ago

what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1

Lostsunrise [7]

Answer:

(n = 7) ⟶ (n = 4)

Explanation:

1. Convert the energy to joules per mole of electrons.

E = 55.1 × 1000 = 55 100 J/mol

2. Convert the energy to joules per electron

E = 55 100/(6.022 × 10²³)

E = 9.150 × 10⁻²⁰ J/electron

3. Use the Rydberg equation to calculate the transition

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes

$\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})$

where

$R_{\text{H}}$ = the Rydberg constant = 2.178 × 10⁻¹⁸ J

$n_{i}$ and $n_{f}$ are the initial and final energy levels.

$9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})$

$\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}$

$n_{f}^{2} = \frac{1 }{ \text{0.062 42}}$

$n_{f}^{2} = 16.02$

$n_{f} = \sqrt{16.02}$

$n_{f} = 4.003 \approx 4$

4 0

3 years ago

NASA pointed the Hubble Space Telescope at a very dark section of sky to see what might be found. The telescope revealed

valentina_108 [34]

Answer:

100,000,000,000

Explanation:

Given that :

1 hubble space telescope pointed at 1 section of the sky reveals 10,000 distinct galaxies ;

Number of sections from which observation is to be made = 10,000,000

Number of distinct galaxy per section = 10,000

Number of galaxies that would exist in the known universe :

Total number of galaxies in the known universe :

Number of galaxies per section * number of sections

10,000 * 10,000,000

= 100,000,000,000s

7 0

3 years ago

Atoms that have lost or gained one or more electrons are called ________.