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soldier1979 [14.2K]

4 years ago

15

A substance made of both metal and nonmetal atoms ?

1 answer:

Nat2105 [25]4 years ago

6 0

A metalloid is a metal and a nonmetal

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Divide velocity by the wavelength.

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4 years ago

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How many digits to the right of the decimal point should be used to report the result?

aalyn [17]

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In engineering and science the common stand is two places.

For example if you get a calculation of 4.567 round up and give the result of 4.57

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3 years ago

Naming ionic compounds what are the structural units that make up ionic compounds and how are they named answers

Marizza181 [45]

Ionic compounds are the combination of two elements, one of which is a metal, while the other is a nonmetal. The intermolecular forces binding them is called an ionic bond. To name an ionic compound, take the name of the metal element first, followed by the nonmetal, but adding the suffix -ide. For example, NaCl is named as sodium chloride.

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3 years ago

Would it be C???...idk<br> HELP

givi [52]

Answer:

C. Butanal , is the aldehyde

Explanation:

A . It is carboxylic acid : ---COOH group

B. It is Ester : ----COOR group , Here R = CH3

C. It is Aldehyde : -----CHO group

D. It is ketone : ----C=O group

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8 0

3 years ago

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

8 0

3 years ago