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2 years ago

does girl code apply for when a girl dates her ex's friend. If not, is it wrong for the girl to do that (15 points)

Chemistry

2 answers:

andreev551 [17]2 years ago

6 0

Answer:

Nope unless u like the ex it’s super wrong but if u like their friend ur good

Explanation:

sleet_krkn [62]2 years ago

4 0

Yes it is wrong bc ur friends it getting over him while she goes off in love with him girls shouorbt do thag

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2 years ago

Methane undergoes combustion. Which products form?

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3 years ago

Which of these is an example of how engineers have affected society?

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3 years ago

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Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

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

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0

3 years ago

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

3 years ago