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3 years ago

HELP ASAPPP!!!!’........

Chemistry

1 answer:

a_sh-v [17]3 years ago

3 0

Answer:

naoh and h2

Explanation:

hope this helps! brainliest please? if not its ok! stay safe and have a good day love <3

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3 years ago

When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?

Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the $\displaystyle ^{202}_{\phantom{2}80}\text{Hg}$ with a neutron $^{1}_{0}n$ . The neutron will add 1 to the mass number 202 of $^{202}_{\phantom{2}80}\text{Hg}$ . However, the atomic number will stay the same.

New mass number: 202 + 1 = 203.
Atomic number is still 80.

$^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}$ .

Double check the equation:

Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The $^{203}_{\phantom{2}80}\text{Hg}$ nucleus loses a proton $^{1}_{1}p$ . Both the mass number 203 and the atomic number will decrease by 1.

New mass number: 203 - 1 = 202.
New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

$^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p$ .

Double check the equation:

Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.

6 0

3 years ago

Help asap please! I award brainliest immediately

grandymaker [24]

Answer: Arsenic has five valence electrons.

Explanation: Good luck!

3 0

3 years ago

Read 2 more answers

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 $V = 2\; \rm dm^{3}$ .

6 0

3 years ago