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3 years ago

What type of rock is this?

Chemistry

2 answers:

777dan777 [17]3 years ago

7 0

Answer: Igneous

Explanation: Igneous rock, or magmatic rock, is one of the three main rock types, the others being sedimentary and metamorphic. Igneous rock is formed through the cooling and solidification of magma or lava. The magma can be derived from partial melts of existing rocks in either a planet's mantle or crust.

zloy xaker [14]3 years ago

6 0

Answer:

I'm gonna say igneous

Explanation:

The texture looks rough and the rock looks like it's composed of different types of minerals making it an igneous rock

I hope this helps :)

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3 years ago

What are the basics of chemistry

Novay_Z [31]

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3 years ago

A hot air balloon is filled to 1250 m3 at 27 C. At what temperature will the balloon be filled to 1600 m3 if the pressure remain

SSSSS [86.1K]

Answer:

384.2 K

Explanation:

First we convert 27 °C to K:

27 °C + 273.16 = 300.16 K

With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:

T₁V₂=T₂V₁

Where in this case:

T₁ = 300.16 K

V₂ = 1600 m³

T₂ = ?

V₁ = 1250 m³

We input the data:

300.16 K * 1600 m³ = T₂ * 1250 m³

And solve for T₂:

T₂ = 384.2 K

7 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$