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3 years ago

Throwing a snow ball during a snow

Chemistry

1 answer:

densk [106]3 years ago

5 0

Answer:

S. Erosion

Explanation:

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tres pensamientos negativos sobre el C0Vid-19 que se vive en la actualidad y luego transformemos a pensamientos positivos.

katrin2010 [14]

Answer:

Bueno,

Explanation:

Tres pesnamientos negativos sobre el C0Vid-19 son,

- Encerrados en la casa

- Tener que ponerse mascara

- Todos los lugares divertidos cerrados

Convertidos en positivos:

- Pasamos mas tiempo de calidad en familia.

- Nos protejemos unos a otros.

- Descansan un poco todas las personas que se han pasado toda la vida trabajando.

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4 years ago

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MgCl2(at)+Br2(l)=MgBr2(at)+Cl2(g)<br><br> how do you write in a word equation

kotykmax [81]

aqueous Magnesium Chloride reacts with liquid Bromide to form aqueous Magnesium Bromide and Chlorine gas

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3 years ago

During a chemical change items are lost or gained to make new substance true or false

Otrada [13]

Answer:

If you meant Atoms then the answer is false

5 0

3 years ago

(chem) which is more concentrated: 45.0 grams of HCOOH dissolved in 189 mL of water or 1.5 moles of CH↓2COOH dissolved in twice

Liula [17]

Answer:

CH3COOH would be more concentrated

Explanation:

The higher the concentration value, the more concentrated it is.

The relationship between concentration, moles and volume is given by the equation;

Concentration = No of moles / Volume

5.0 grams of HCOOH dissolved in 189 mL of water

Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol

Concentration = 0.1086 / 0.189 = 0.5746 mol/L

1.5 moles of CH3COOH dissolved in twice as much water

Volume = 2 * 189 = 378 ml = 0.378 L

Concentration = 1.5 / 0.378 = 3.9683 mol/L

Comparing both concentration values;

CH3COOH would be more concentrated

6 0

3 years ago

Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife

yuradex [85]

Answer:

$\boxed{\text{2408 min}}$

Explanation:

The integrated rate law for radioactive decay is

$\ln\dfrac{N_{0}}{N_{t}} = kt$

1. Calculate the decay constant

$\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}$

8 0

3 years ago