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IgorC [24]
3 years ago
8

What is the correct name for the compound with the formula CrPO4 and why

Chemistry
2 answers:
Alla [95]3 years ago
8 0
B djdjjdjddjdjdjdidjuhdbshshxbhxdhdjdxx. It is B
MAVERICK [17]3 years ago
6 0

Answer:

The answer is D. Chromium (|||) Phosphate

Explanation:

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How much energy is needed to warm 7.40g of water by 55 degress C?
zloy xaker [14]
<span>E = mCdT
E = energy, m = mass, C = specific heat capacity, dT = change in temperature.
526 = 0.074C x 17
E = 0.074C x 55
Divide the equations
E/526 = (0.074C x 55)/(0.074C x 17) = 55/17
E = (55 x 526)/17 = 1702 J</span>
8 0
3 years ago
Please help me :( Thank you so much ❤️
m_a_m_a [10]

Answer:

G- Gallons-Miles

Explanation

Even though gallons of gas are converted to miles you cannot physically convert gallons of something to miles.

6 0
3 years ago
Matter of the universe is made of building blocks known as atoms TrueFalse
zaharov [31]

True. The building blocks of life are atoms

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3 years ago
What Type of Reaction is the following: nickel+ fluorine → nickel (II) fluoride
Sphinxa [80]

Answer:

c

Explanation:

7 0
3 years ago
A sample of an unknown gas takes 371 s to diffuse through a porous plug at a given temperature.
blsea [12.9K]

Answer:

125.84 g/mol is the molar mass of the unknown gas.

Explanation:

Let the volume of the gases effusing out be V.

Effusion rate of the unknown gas = R=\frac{V}{371 s}

Effusion rate of the nitrogen gas = r=\frac{V}{175 s}

Molar mass of unknown gas = m

Mass of nitrogen gas = 28 g/mol

Graham's law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{R}{r}=\sqrt{\frac{28 g/mol}{M}}

\frac{\frac{V}{371 s}}{\frac{V}{175s}}=\sqrt{\frac{28 g/mol}{M}}

\frac{175 s}{371 s}=\sqrt{\frac{28 g/mol}{M}}

M=\frac{28 g/mol\times 371\times 371}{175\times 175}=125.84 g/mol

125.84 g/mol is the molar mass of the unknown gas.

5 0
3 years ago
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