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2 years ago

Hi :) take some points for the road

Chemistry

2 answers:

Maurinko [17]2 years ago

6 0

Answer:

Explanation:

zhuklara [117]2 years ago

3 0

Answer:

oh rally? thank you >//<

Explanation:

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What is the correct name for ionic compound Cs2S?

Leno4ka [110]

The name for the ionic compound Cs2S is Cesium sulfide.

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2 years ago

Is glass biotic or abiotic

zhuklara [117]

Glass is not a living organism, so it is <u>abiotic</u>.

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2 years ago

Read 2 more answers

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

1 year ago

Chemosynthesis and photosynthesis are both processes that produce food. <br><br> True or false?

MA_775_DIABLO [31]

Answer:

true

Explanation:

probably true

8 0

2 years ago

A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

koban [17]

Answer:

$Ca^{2+} and Co^{2+}$

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, $Ca_2$ and $Co_2$

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding $Li_2SO_4$ to it precipitation takes place.

Precipitate is as follows,

$Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)$

Thus,

$Ca^2^+$ is present

When $Li_3PO_4$ is added again precipitation takes place.

Therefore the reaction is as follows,

$Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)$

Therefore,

$Ca^{2+} and Co^{2+}$ are present in the solution

3 0

2 years ago