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2 years ago

Is anyone good at high school chemistry ? If so can you help me please

Chemistry

1 answer:

PolarNik [594]2 years ago

7 0

Answer:

2.25?

Explanation:

Im a 7th grade and have no idea but I think the pressure Will be 2.25 ATM because 3 which is the volume after minus the 0.75 which was the first volume but I dont really know srry

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The maximum number of electrons in a p sub-level is:

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How many grams of MgO are needed to produce 264.0 grams of Mg(OH)2?

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Which of the following is the correctly balanced chemical equation for the reaction of Ca(OH)2 and HNO3?

sattari [20]

Explanation:

The reaction between calcium hydroxide and nitric acid is as follows.

$Ca(OH)_{2} + HNO_{3} \rightarrow Ca(NO_{3})_{2} + H_{2}O$

Number of reactant atoms are as follows.

Ca = 1
O = 4
H = 3
N = 1

Number of product atoms are as follows.

Ca = 1
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H = 2
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To balance the given chemical equation, multiply $HNO_{3}$ by 2 on reactant side and multiply $H_{2}O$ by 2 on the product side.

Therefore, the balanced chemical equation will be as follows.

$Ca(OH)_{2} + 2HNO_{3} \rightarrow Ca(NO_{3})_{2} + 2H_{2}O$

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3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

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3 years ago

Solutions that are very concentrated have greater freezing point depression Group of answer choices true or false

Tamiku [17]

Answer:

Explanation:

False. The greater the concentration, the lower the freezing point.

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