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Marta_Voda [28]

3 years ago

14

All my questions are being "Deleted" for violating community guidelines by the same bots that scam you with the file. I swear- i

f it turns out the moderation team is scamming it's own site users.... (Not saying this is the case, but how does a scam 'bot' delete a question?)

1 answer:

Goryan [66]3 years ago

3 0

Answer:

well im not sure why

Explanation:

but this one isn't deleted yet sooo

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Which solution below yields the lowest final concentration of a sodium hydroxide solution prepared by diluting 50.00 mL of conce

Firdavs [7]

Answer:

The answer to your question is: letter D. 1.33 L

Explanation:

Data

V1 = 50 ml

C1 = 19.3

To solve this problem use the formula C₁V₁ = C₂V₂

C2 = C1V1 / V2

C = concentration

V = volume

a) 1.15 L

C2 = (19.3)(50) / 1150

C2 = 0.84 M

b) No right answer

c) V2= 0.80 L

C2 = (19.3)(50) / 800

C2 = 1.2 M

d) V2 = 1.33 L

C2 = (19.3)(50) / 1330

C2 = 0.72 M

e) V2 = 350 ml

C2 = (19.3)(50) / 350

C2 = 2.75 M

3 0

3 years ago

URGENT!! Can someone please explain this to me?

djverab [1.8K]

Answer:

$CH _{4}+ 2O _{2} → CO _{2} + 2H _{2}O$

for the balanced equation

$from \: the \: equation \\ 1 \: mole \: of \: methane \: gives \: 1 \: mole \: of \: carbon \: dioxide \\ 7.4 \: moles \: of \: carbon \: dioxide \: will \: be \: given \: by \: (7.4 \times 1)moles \: of \: methane \\ = 7.4\: moles \: of \: methane \\ \\ since \: moles \: of \: oxygen \: double \: those \: of \: methane \\ moles \: of \: oxygen \: = 7.4\times 2 \\ = 14.8 \: moles \: of \: oxygen$

4 0

3 years ago

The elements in Groups 1A(1) and 7A(17) are all quite reactive. What is a major difference between them?

matrenka [14]

The elements in Groups 1A(1) and 7A(17) are all quite reactive.

<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>

Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.

Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.

Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
Reactivity increase down group 1 but decrease up group 7 this is because group 7 elements react by gaining an electron. As one move down the group, the amount of electron shielding increases, meaning that the electron is less attracted to the nucleus.

To know more about Groups 1A(1) and 7A(17) please click here :

brainly.com/question/13063502

#SPJ4

6 0

2 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

3 years ago

1 When a chemical reaction occurs A. the substances involved do not mix together and maintain their original properties. B. the

nlexa [21]

Answer:

D

Explanation:

The proberties of the substances that are produced are different from the properties of the original substances.

6 0

3 years ago