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3 years ago

All bases dissociate True or false

Chemistry

1 answer:

Nesterboy [21]3 years ago

6 0

Answer:

verdadero

Explanation:

porque esoo $\lim_{n \to \infty} a_n x_{123} \frac{x}{y} \sqrt[n]{x} x^{2} \sqrt{x} \pi \neq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right.$

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Contact [7]

To cause death within hours of exposure to radiation, the dose needs to be very high, 10Gy or higher, while 4-5Gy will kill within 60 days, and less than 1.5-2Gy will not be lethal in the short term. However all doses, no matter how small, carry a finite risk of cancer and other diseases. Patients exposed to radiation between 8 to 30 Gy experience nausea and severe diarrhea within an hour, and they die between 2 days and 2 weeks after exposure. Absorbed doses greater than 30 Gy cause neurological damage

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3 years ago

Thermal energy _____ as temperature increases

Romashka-Z-Leto [24]

A. increases

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3 years ago

A sample of xenon gas at a pressure of 1.14 atm and a temperature of 20.4 °C, occupies a volume of 16.9 liters. If the gas is al

kondor19780726 [428]

Answer:

P2= 0.696atm

Explanation:

Applying Boyle's law

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3 years ago

ANSWER ASAP (20 POINTS)

SIZIF [17.4K]

Answer:

a chemical formula

Explanation:

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3 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0

4 years ago