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Stells [14]
3 years ago
12

Why must we do the a lot of quantity urine​

Chemistry
1 answer:
serg [7]2 years ago
7 0

Answer:

because

ExplanatioN:

<em>BeCaUsE</em>

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Determine the molecular formula of a compound that has a molar mass of 183.2 g/mol and an empirical formula of c2h5o2.
Tpy6a [65]
C6H15O6

Good luck and don't forget to rate or mark Brainliest :)
6 0
3 years ago
Read 2 more answers
Can someone help pls
zysi [14]
What is the problem u need help with
3 0
2 years ago
2. The pressure of the oxygen gas inside a
Flauer [41]

Answer: 4.41 atm

Explanation:

Given that,

Original pressure of oxygen gas (P1) = 5.00 atm

Original temperature of oxygen gas (T1) = 25°C

[Convert 25°C to Kelvin by adding 273

25°C + 273 = 298K

New pressure of oxygen gas (P2) = ?

New temperature of oxygen gas (T2) = -10°C

[Convert -10°C to Kelvin by adding 273

-10°C + 273 = 263K

Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law

P1/T1 = P2/T2

5.00 atm /298K = P2/263K

To get the value of P2, cross multiply

5.00 atm x 263K = 298K x V2

1315 atm•K = 298K•V2

V2 = 1315 atm•K / 298K

V2 = 4.41 atm

Thus, the new pressure inside the canister is 4.41 atmosphere

4 0
3 years ago
Which element or elements are unbalanced in this equation?
Stells [14]

Answer:

Only the H is unbalanced.

Explanation:

There are 4 H's and 2 of everything else

7 0
3 years ago
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An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
topjm [15]

Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

Data:

n_{f} = 2

λ = 657 nm

Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

7 0
3 years ago
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